GIFT  OF 
^ 

^^*A^4 


MATHEMATICAL   TEXT- BOOKS 
By  G.  A.  WENTWORTH,  A.M. 

Mental  Arithmetic. 

Elementary  Arithmetic. 

Practical  Arithmetic. 

Primary  Arithmetic. 

Grammar  School  Arithmetic. 

High  School  Arithmetic. 

High  School  Arithmetic  (Abridged). 

First  Steps  in  Algebra. 

School  Algebra. 

College  Algebra. 

Elements  of  Algebra. 

Complete  Algebra. 

Shorter  Course  in  Algebra. 

Higher  Algebra. 

New  Plane  Geometry. 

New  Plane  and  Solid  Geometry. 

Syllabus  of  Geometry. 

Geometrical  Exercises. 

Plane  and  Solid  Geometry  and  Plane  Trigonometry. 

New  Plane  Trigonometry. 

New  Plane  Trigonometry,  with  Tables. 

New  Plane  and  Spherical  Trigonometry. 

New  Plane  and  Spherical  Trig.,  with  Tables. 

New  Plane  and  Spherical  Trig.,  Surv.,  and  Nav. 

New  Plane  Trig,  and  Surv.,  with  Tables. 

New  Plane  and  Spherical  Trig.,  Surv.,  with  Tables. 

Analytic  Geometry. 


TEXT-BOOK 


Off 


GEOMETEY 


REVISED    EDITION. 


BY 

G.  A.  WENTWORTH,  A.M., 

AUTHOR   OF  A   SERIES   OF  TEXT-BOOKS   IN   MATHEMATICS. 


BOSTON,  U.S.A.: 
PUBLISHED   BY   GINN   &   COMPANY. 


v\)3 


Entered,  according  to  Act  of  Congress,  in  the  year  1888,  by 

G.  A.  WENT  WORTH, 
in  the  Office  ot  the  Librarian  of  Congress,  at  Washington. 


ALL  RIGHTS  RESERVED. 


' 


TYPOGRAPHY  BY  J.  S.  GUSHING  &  Co.,  BOSTON,  U.S.A. 


PBESSWORK  BY  GINN  &  Co.,  BOSTON,  U.S.A. 


PREFACE. 


"TV  yf~OST  persons  do  not  possess,  and  do  not  easily  acquire,  the  power 
•"-•-  of  abstraction  requisite  for  apprehending  geometrical  concep- 
tions, and  for  keeping  in  mind  the  successive  steps  of  a  continuous 
argument.  Hence,  with  a  very  large  proportion  of  beginners  in  Geom- 
etry, it  depends  mainly  upon  the  form  in  which  the  subject  is  pre- 
sented whether  they  pursue  the  study  with  indifference,  not  to  say 
aversion,  or  with  increasing  interest  and  pleasure. 

In  compiling  the  present  treatise,  the  author  has  kept  this  fact  con- 
stantly in  view.  All  unnecessary  discussions  and  scholia  have  been 
avoided ;  and  such  methods  have  been  adopted  as  experience  and 
attentive  observation,  combined  with  repeated  trials,  have  shown  to  be 
most  readily  comprehended.  No  attempt  has  been  made  to  render 
more  intelligible  the  simple  notions  of  position,  magnitude,  and  direc- 
tion, which  every  child  derives  from  observation  ;  but  it  is  believed 
that  these  notions  have  been  limited  and  defined  with  mathematical 
precision. 

A  few  symbols,  which  stand  for  words  and  not  for  operations,  have 
been  used,  but  these  are  of  so  great  utility  in  giving  style  and  per- 
spicuity to  the  demonstrations  that  no  apology  seems  necessary  for 
their  introduction. 

Great  pains  have  been  taken  to  make  the  page  attractive.  The 
figures  are  large  and  distinct,  and  are  placed  in  the  middle  of  the 
page,  so  that  they  fall  directly  under  the  eye  in  immediate  connec- 
tion with  the  corresponding  text.  The  given  lines  of  the  figures  are 
full  lines,  the  lines  employed  as  aids  in  the  demonstrations  are  short- 
dotted,  and  the  resulting  lines  are  long-dotted. 

327374 


iv  PREFACE. 

In  each  proposition  a  concise  statement  of  what  is  given  is  printed 
in  one  kind  of  type,  of  what  is  required  in  another,  and  the  demon- 
stration in  still  another.  The  reason  for  each  step  is  indicated  in 
small  type  between  that  step  and  the  one  following,  thus  preventing 
the  necessity  of  interrupting  the  process  of  the  argument  by  referring 
to  a  previous  section.  The  number  of  the  section,  however,  on  which 
the  reason  depends  is  placed  at  the  side  of  the  page.  The  constituent 
parts  of  the  propositions  are  carefully  marked.  Moreover,  each  distinct 
assertion  in  the  demonstrations:  and  each  particular  direction  in  the 
construction  of  the  figures,  begins  a  new  line;  and  in  no  case  is  it 
necessary  to  turn  the  page  in  reading  a  demonstration. 

This  arrangement  presents  obvious  advantages.  The  pupil  perceives 
at  once  what  is  given  and  what  is  required,  readily  refers  to  the  figure 
at  every  step,  becomes  perfectly  familiar  with  the  language  of  Geom- 
etry, acquires  facility  in  simple  and  accurate  expression,  rapidly  learns 
to  reason,  and  lays  a  foundation  for  completely  establishing  the 
science. 

Original  exercises  have  been  given,  not  so  difficult  as  to  discourage 
the  beginner,  but  well  adapted  to  afford  an  effectual  test  of  the  degree 
in  which  he  is  mastering  the  subjects  of  his  reading.  Some  of  these 
exercises  have  been  placed  in  the  early  part  of  the  work  in  order 
that  the  student  may  discover,  at  the  outset,  that  to  commit  to  mem- 
ory a  number  of  theorems  and  to  reproduce  them  in  an  examination 
is  a  useless  and  pernicious  labor ;  but  to  learn  their  uses  and  appli- 
cations, and  to  acquire  a  readiness  in  exemplifying  their  utility  is  to 
derive  the  full  benefit  of  that  mathematical  training  which  looks  not 
BO  much  to  the  attainment  of  information  as  to  the  discipline  of  the 
mental  faculties. 

G.  A.  WENTWORTH. 
EXETER,  N.H. 
1878. 


PREFACE. 


TO   THE   TEACHER. 

WHEN  the  pupil  is  reading  each  Book  for  the  first  time,  it  will  be 
well  to  let  him  write  his  proofs  on  the  blackboard  in  his  own  lan- 
guage ;  care  being  taken  that  his  language  be  the  simplest  possible, 
that  the  arrangement  of  work  be  vertical  (without  side  work),  and 
that  the  figures  be  accurately  constructed. 

This  method  will  furnish  a  valuable  exercise  as  a  language  lesson, 
will  cultivate  the  habit  of  neat  and  orderly  arrangement  of  work, 
and  will  allow  a  brief  interval  for  deliberating  on  each  step. 

After  a  Book  has  been  read  in  this  way,  the  pupil  should  review 
the  Book,  and  should  be  required  to  draw  the  figures  free-hand.  He 
should  state  and  prove  the  propositions  orally,  using  a  pointer  to 
indicate  on  the  figure  every  line  and  angle  named.  He  should  be 
encouraged,  in  reviewing  each  Book,  to  do  the  original  exercises  ;  to 
state  the  converse  of  propositions ;  to  determine  from  the  statement, 
if  possible,  whether  the  converse  be  true  or  false,  and  if  the  converse 
be  true  to  demonstrate  it ;  and  also  to  give  well-considered  answers 
to  questions  which  may  bet  asked  him  on  many  propositions. 

The  Teacher  is  strongly  advised  to  illustrate,  geometrically  and 
arithmetically,  the  principles  of  limits.  Thus  a  rectangle  with  a  con- 
stant base  6,  and  a  variable  altitude  x,  will  afford  an  obvious  illus- 
tration of  the  axiomatic  truth  that  the  product  of  a  constant  and  a 
variable  is  also  a  variable  ;  and  that  the  limit  of  the  product  of  a 
constant  and  a  variable  is  the  product  of  tbe  constant  by  the  limit 
of  the  variable.  If  x  increases  and  approaches  the  altitude  a  as  a 
limit,  the  area  of  the  rectangle  increases  and  approaches  the  area  of 
the  rectangle  ab  as  a  limit;  if,  however,  x  decreases  and  approaches 
zero  as  a  limit,  the  area  of  the  rectangle  decreases  and  approaches 
zero  for  a  limit.  An  arithmetical  illustration  of  this  truth  may  be 
given  by  multiplying  a  constant  into  the  approximate  values  of  any 
repetend.  If,  for  example,  we  take  the  constant  60  and  the  repetend 
0.3333,  etc.,  the  approximate  values  of  the  repetend  will  be  T%,  T%3<j-, 


VI  PREFACE. 


^,  etc.,  and  these  values  multiplied  by  60  give  the  series 
18,  19.8,  19.98,  19.998,  etc.,  which  evidently  approaches  20  as  a  limit; 
but  the  product  of  60  into  J  (the  limit  of  the  repetend  0.333,  etc.)  is 
also  20. 

Again,  if  we  multiply  60  into  the  different  values  of  the  decreasing 
series  ^,  7^7,  3^5-,  j-sfav'  e^c.,  which  approaches  zero  as  a  limit,  we 
shall  get  the  decreasing  series  2,  £,  ^,  ?J7,  etc.;  and  this  series  evi- 
dently approaches  zero  as  a  limit. 

In  this  way  the  pupil  may  easily  be  led  to  a  complete  compre- 
hension of  the  subject  of  limits. 

The  Teacher  is  likewise  advised  to  give  frequent  written  examina- 
tions. These  should  not  be  too  difficult,  and  sufficient  time  should  be 
allowed  for  accurately  constructing  the  figures,  for  choosing  the  best 
language,  and  for  determining  the  best  arrangement. 

The  time  necessary  for  the  reading  of  examination-books  will  be 
diminished  by  more  than  one-half,  if  the  use  of  the  symbols  employed 
in  this  book  be  allowed. 

G.  A.  W. 
EXETER,  N.H. 

1879. 


PKEFACE.  Vll 


NOTE   TO   REVISED   EDITION. 

THE  first  edition  of  this  Geometry  was  issued  about  nine  years  ago. 
The  book  was  received  with  such  general  favor  that  it  has  been  neces- 
sary to  print  very  large  editions  every  year  since,  so  that  the  plates 
are  practically  worn  out.  Taking  advantage  of  the  necessity  for  new 
plates,  the  author  has  re-written  the  whole  work  ;  but  has  retained 
all  the  distinguishing  characteristics  of  the  former  edition.  A  few 
changes  in  the  order  of  the  subject-matter  have  been  made,  some  of 
the  demonstrations  have  been  given  in  a  more  concise  and  simple 
form  than  before,  and  the  treatment  of  Limits  and  of  Loci  has  been 
made  as  easy  of  comprehension  as  possible. 

More  than  seven  hundred  exercises  have  been  introduced  into  this 
edition.  These  exercises  consist  of  theorems,  loci,  problems  of  con- 
struction, and  problems  of  computation,  carefully  graded  and  specially 
adapted  to  beginners.  No  geometry  can  now  receive  favor  unless  it 
provides  exercises  for  independent  investigation,  which  must  be  of  such 
a  kind  as  to  interest  the  student  as  soon  as  he  becomes  acquainted 
with  the  methods  and  the  spirit  of  geometrical  reasoning.  The  author 
has  observed  with  the  greatest  satisfaction  the  rapid  growth  of  the 
demand  for  original  exercises,  and  he  invites  particular  attention  to 
the  systematic  and  progressive  series  of  exercises  in  this  edition. 

The  part  on  Solid  Geometry  has  been  treated  with  much  greater 
freedom  than  before,  and  the  formal  statement  of  the  reasons  for  the 
separate  steps  has  fegfen  in  general  omitted,  for  the  purpose  of  giving  a 
more  elegant  form  tt>  the  demonstrations. 

A  brief  treatise  dn  Conic  Sections  (Book  IX)  has  been  prepared, 
and  is  issued  in  pamphlet  form,  at  a  very  low  price.  It  will  also  be 
bound  with  the  Geometry  if  that  arrangement  is  found  to  be  gen- 
erally desired. 


vili  PREFACE. 

The  author  takes  this  opportunity  to  express  his  grateful  appre- 
ciation of  the  generous  reception  given  to  the  Geometry  heretofore  by 
the  great  body  of  teachers  throughout  the  country,  and  he  confidently 
anticipates  the  same  generous  judgment  of  his  efforts  to  bring  the  work 
up  to  the  standard  required  by  the  great  advance  of  late  in  the  sci- 
ence and  method  of  teaching. 

The  author  is  indebted  to  many  correspondents  for  valuable  sug- 
gestions ;  and  a  special  acknowledgment  is  due,  for  criticisms  and 
careful  reading  of  proofs,  to  Messrs.  C.  H.  Judson,  of  Greenville,  S.C. ; 
Samuel  Hart,  of  Hartford,  Conn. ;  J.  M.  Taylor,  of  Hamilton,  N.Y. ; 
W.  Le  Conte  Stevens,  of  Brooklyn,  N.Y. ;  E.  R.  Offutt,  of  St.  Louis, 
Mo.;  J.  L.  Patterson,  of  Lawrenceville,  N.  J.;  G.  A.  Hill,  of  Cam- 
bridge, Mass. ;  T.  M.  Blakslee,  of  Des  Moines,  la.;  G.  W.  Sawin,  of  Cam- 
bridge, Mass. ;  Ira  M.  De  Long,  of  Boulder,  Col. ;  and  W.  J.  Lloyd,  of 
New  York,  N.Y. 

Corrections  or  suggestions  will  be  thankfully  received. 

G.  A.  WENTWORTH. 
EXETER,  N.H., 
1888. 


CONTENTS. 


GEOMETRY. 

PAGE 

DEFINITIONS 1 

STRAIGHT  LINES 5 

PLANE  ANGLES 7 

MAGNITUDE  OF  ANGLES 9 

ANGULAR  UNITS 10 

METHOD  OF  SUPERPOSITION 11 

SYMMETRY 13 

MATHEMATICAL  TERMS 14 

POSTULATES 15 

AXIOMS    .        . .16 

SYMBOLS  .               16 


PLANE   GEOMETRY. 

BOOK  I.     THE  STRAIGHT  LINE. 

THE  STRAIGHT  LINE 17 

PARALLEL  LINES 22 

PERPENDICULAR  AND  OBLIQUE  LINES     .....  33 

TRIANGLES 40 

QUADRILATERALS .        .  56 

POLYGONS  IN  GENERAL 66 

EXERCISES        , 72 


X  CONTENTS. 

BOOK  II.    THE  CIRCLE. 

PAGE 

DEFINITIONS 75 

AECS  AND  CHORDS   .        . 77 

TANGENTS 89 

MEASUREMENT.                92 

THEORY  OF  LIMITS 94 

MEASURE  OF  ANGLES 98 

PROBLEMS  OF  CONSTRUCTION 106 

EXERCISES 126 

BOOK  III.   PROPORTIONAL  LINES  AND  SIMILAR  POLYGONS. 

THEORY  OF  PROPORTION 131 

PROPORTIONAL  LINES 138 

SIMILAR  TRIANGLES .        .  145 

SIMILAR  POLYGONS 153 

NUMERICAL  PROPERTIES  OF  FIGURES 156 

PROBLEMS  OF  CONSTRUCTION 167 

PROBLEMS  OF  COMPUTATION 173 

EXERCISES 175 

BOOK  IV.    AREAS  OF  POLYGONS. 

AREAS  OF  POLYGONS .  180 

COMPARISON  OF  POLYGONS 188 

PROBLEMS  OF  CONSTRUCTION 192 

PROBLEMS  OF  COMPUTATION 204 

EXERCISES ,  205 

BOOK  V.     REGULAR  POLYGONS  AND  CIRCLES. 

REGULAR  POLYGONS  AND  CIRCLES 209 

PROBLEMS  OF  CONSTRUCTION 222 

MAXIMA  AND  MINIMA 230 

EXERCISES 237 

MISCELLANEOUS  EXERCISES     ....                       .  240 


GEOMETRY. 


DEFINITIONS. 

1,  If  a  block  of  wood  or  stone  be  cut  in  the  shape  repre- 
sented in  Fig.  1,  it  will  have  six  flat  faces. 

Each  face  of  the  block  is  called 
a  surface ;  and  if  these  faces  are  made  **f 
smooth  by  polishing,  so  that,  when 
a  straight-edge  is  applied  to  any  one 
of  them,  the  straight  edge  in  every 
part  will  touch  the  surface,  the  faces 
are  called  plane  surfaces,  or  planes. 

2,  The  edge  in  which  any  two  of  these  surfaces  meet  is 
called  a  line. 

3,  The  corner  at  which  any  three  of  these  lines  meet  is 
called  a  point. 

4,  For  computing  its  volume,  the  block  is  measured  in  three 
principal  directions : 

From  left  to  right,  A  to  B. 
From  front  to  back,  A  to  C. 
From  bottom  to  top,  A  to  D. 

These  three  measurements  are  called  the  dimensions  of  the 
block,  and  are  named  length,  breadth  (or  width),  thickness 
(height  or  depth). 


•GEOMETRY. 


A  solid,  therefore,  has  three  dimensions,  length,  breadth,  and 
thickness. 

5,  The  surface  of  a  solid  is  no  part  of  the  solid.      It  is 
simply  the  boundary  or  limit  of  the  solid.     A  surface,  there- 
fore, has  only  two  dimensions,  length  and  breadth.      So  that, 
if  any  number  of  flat   surfaces  be  put  together,    they  will 
coincide  and  form  one  surface. 

6,  A  line  is  no  part  of  a  surface.     It  is  simply  a  boundary 
or  limit  of  the  surface.     A  line,  therefore,  has  only  one  dimen- 
sion, length.     So  that,  if  any  number  of  straight  lines  be  put 
together,  they  will  coincide  and  form  one  line. 

7,  A  point  is  no  part  of  a  line.      It  is  simply  the  limit  of 
the  line.     A  point,   therefore,   has  no  dimension,   but  denotes 
position  simply.     So  that,   if  any  number  of  points  be  put 
together,  they  will  coincide  and  form  a  single  point. 

8,  A  solid,   in  common  language,  is  a  limited  portion  of 
space  filled  with  matter ;  but  in  Geometry  we  have  nothing 
to  do  with  the  matter  of  which  a  body  is  composed  ;  we  study 
simply  its  shape  and  size;  that  is,   we  regard  a  solid  as  a 
limited  portion  of  space  which  may  be -occupied  by  a  physical 
body,  or  marked  out  in  some  other  way.     Hence, 

A  geometrical  solid  is  a  limited  portion  of  space. 

9,  It  must  be  distinctly  understood  at  the  outset  that  the 
points,   lines,   surfaces,   and  solids  of  Geometry  are  purely 
ideal,  though  they  can  be  represented  to  the  eye  in  only  a 
material  way.      Lines,  for  example,  drawn  on  paper  or  on  the 
blackboard,  will  have  some  width  and  some  thickness,  and 
will  so  far  fail  of  being  true  lines ;  yet,  when  they  are  used  to 
help  the  mind  in  reasoning,  it  is  assumed  that  they  represent 
perfect  lines,  without  breadth  and  without  thickness. 


DEFINITIONS.  3 

10,  A  point  is  represented  to  the  eye  by  a  fine  dot,  and 
named  by  a  letter,  as  A  (Fig.  2) ;  a  line  is  named  by  two 
letters,   placed  one  at  each  end, 

as  BF\  a  surface  is  represented 
and  named  by  the  lines  which 
bound  it,  as  BCDF;  a  solid  is 
represented  by  the  faces  which 
bound  it.  FlQ-  2- 

11,  By  supposing  a  solid  to  diminish  gradually  until  it 
vanishes  we  may  consider  the  vanishing  point,   a  point  in 
space,  independent  of  a  line,  having  position  but  no  extent. 

12,  If  a  point  moves  continuously  in  space,  its  path  is  a 
line.     This  line  may  be  supposed  to  be  of  unlimited  extent, 
and  may  be  considered  independent  of  the  idea  of  a  surface. 

13,  A  surface  may  be  conceived  as  generated  by  a  line 
moving  in  space,  and  as  of  unlimited  extent.     A  surface  can 
then  be  considered  independent  of  the  idea  of  a  solid. 

14,  A  solid  may  be  conceived  as  generated  by  a  surface  in 
motion. 

Thus,  in  the  diagram,  let  the  up-         D  ^r 

right  surface   ABCD  move   to  the        I  ~"K? 

right  to  the  position  EFGH.     The    Ap- 
points A,  B,  C,  and  D  will  generate  < 
the  lines  AE,  BF,   CG,  and  DH,        j_  ~~\/_ 
respectively.     The   lines   AB,    BO,                 ^ 
CD,  and  AD  will  generate  the  sur- 
faces AF,   BG,    OH,   and  AH,  respectively.     The  surface 
ABCD  will  generate  the  solid  AG. 

15,  Geometry  is  the  science  which  treats  of  position,  form, 
and  magnitude. 

16,  Points,  lines,  surfaces,  and  solids,  with  their  relations, 
constitute  the  subject-matter  of  Geometry. 


G 
'F 


4  GEOMETRY. 

17,  A  straight  line,  or  right  line,  is  a  line  which  has  the 

same  direction  throughout  its     » g 

whole  extent,  as  the  line  AB. 

18,  A  curved  line  is  a  line 
no  part  of  which  is  straight, 
as  the  line  CD. 

19,  A  broken  line  is  a  series 
of  different  successive  straight 

lines,  as  the  line  EF.  Fm-  4' 

20,  A  mixed  line  is  a  line  composed  of  straight  and  curved 
lines,  as  the  line  GH. 

A  straight  line  is  often  called  simply  a  line,  and  a  curved 
line,  a  curve. 

21,  A  plane  surface,  or  a  plane,  is  a  surface  in  which,  if 
any  two  points  be  taken,  the  straight  line  joining  these  points 
will  lie  wholly  in  the  surface. 

22,  A  curved  surface  is  a  surface  no  part  of  which  is  plane. 

23,  Figure  or  form  depends  upon  the  relative  position  of 
points.     Thus,  the  figure  or  form  of  a  line  (straight  or  curved) 
depends  upon  the  relative  position  of  the  points  in  that  line  ; 
the  figure  or  form  of  a  surface  depends  upon  the  relative 
position  of  the  points  in  that  surface. 

24,  With  reference  to  form  or  shape,  lines,  surfaces,  and 
solids  are  called  figures. 

With  reference  to   extent,   lines,   surfaces,  and   solids  are 
called  magnitudes. 

25,  A  plane  figure  is  a  figure  all  points  of  which  are  in  the 
same  plane. 

26,  Plane  figures  formed  by  straight  lines  are  called  rec- 
tilinear  figures ;    those    formed   by   curved  lines   are   called 
curvilinear  figures  ;  and  those  formed  by  straight  and  curved 
lines  are  called  mixtilinear  figures. 


DEFINITIONS.  5 

27,  Figures  which  have  the  same  shape  are  called  similar 
figures.      Figures  which  have  the  same  size  are  called  equiva- 
lent figures.     Figures  which  have  the  same  shape  and  size  are 
called  equal  or  congruent  figures. 

28,  Geometry  is  divided  into  two  parts,  Plane  Geometry 
and  Solid  Geometry.      Plane  Geometry  treats  of  figures  all 
points  of  which  are    in    the  same  plane.     Solid   Geometry 
treats  of  figures  all  points  of  which  are  not  in  the  same  plane. 

STRAIGHT  LINES. 

29,  Through  a  point  an  indefinite  number  of  straight  lines 
may  be  drawn.     These  lines  will  have  different  directions. 

30,  If  the  direction  of  a  straight  line  and  a  point  in  the 
line  are  known,  the  position  of  the  line  is  known  ;  in  other 
words,  a  straight  line  is  determined  if  its  direction  and  one  of 
its  points  are  known.     Hence, 

All  straight  lines  which  pass  through  the  same  point  in  the 
same  direction  coincide,  and  form  but  one  line. 

31,  Between  two  points  one,  and  only  one,  straight  line 
can  be  drawn  ;  in  other  words,  a  straight  line  is  determined 
if  two  of  the  points  are  known.     Hence, 

Two  straight  lines  which  have  two  points  in  common  coincide 
throughout  their  whole  extent,  and  form  but  one  line. 

32,  Two  straight  lines  can  intersect  (cut  each  other)  in  only 
one  point ;  for  if  they  had  two  points  common,  they  would 
coincide  and  not  intersect. 

33,  Of  all  lines  joining  two  points  the  shortest  is  the  straight 
line,  and  the  length  of  the  straight  line  is  called  the  distance 
between  the  two  points. 


0  GEOMETRY. 

34,  A  straight  line  determined  by  two  points  is  considered 
as  prolonged  indefinitely  both  ways.     Such  a  line  is  called  an 
indefinite  straight  line. 

35,  Often  only  the  part  of  the  line  between  two  fixed  points 
is  considered.     This  part  is  then  called  a  segment  of  the  line. 

For  brevity,  we  say  "the  line  AB"  to  designate  a  segment 
of  a  line  limited  by  the  points  A  and  B. 

36,  Sometimes,  also,  a  line  is  considered  as  proceeding  from 
a  fixed  point  and  extending  in  only  one  direction.     This  fixed 
point  is  then  called  the  origin  of  the  line. 

37,  If  any  point  C  be  taken  in  a  given  straight  line  AB,  the 
two  parts  CA  and  CB  aro 

said  to  have  opposite  direc-     A  -  ±  -  B 
tions  from  the  point  C.  Fm.  5. 

38,  Every  straight  line,  as  AB,  may  be  considered  as  hav- 
ing opposite  directions,  namely,  from  A  towards  B,  which  is 
expressed  by  saying  "line  AB"\  and  from  B  towards^,  which 
is  expressed  by  saying  "line  BA" 

39,  If  the  magnitude  of  a  given  line  is  changed,  it  becomes 
longer  or  shorter. 

Thus  (Fig.  5),  by  prolonging  AC  to  B  we  add  CB  to  AC, 
and  AB  —  AC+  CB.  By  diminishing  AB  to  C,  we  subtract 
CB  from  AB,  and  AC=  AB  -  CB. 

If  a  given  line  increases  so  that  it  is  prolonged  by  its  own 
magnitude  several  times  in  ^  ^  _  D  E 

succession,  the  line  is  multi-      H  -  «  -  '  -  '  -  h 
plied,  and  the  resulting  line 

is  called  a  multiple  of  the  given   line.      Thus   (Fig.   6),    if 

AC=2AB,  AD  =  SAB,  and 


Hence, 


DEFINITIONS. 


Lines  of  given  length  may  be  added  and  subtracted;   they 
may  also  be  multiplied  and  divided  by  a  number. 


FIG. 


PLANE  ANGLES. 

40,  The  opening  between  two  straight  lines  which  meet  is 
called  a  plane  angle.  The  two  lines  are  called  the  sides,  and 
the  point  of  meeting,  the  vertex,  of  the  angle. 


41.  If  there  is  but  one  angle  at  a 
given  vertex,  it  is  designated  by  a  cap- 
ital letter  placed  at  the  vertex,  and  is 
read  by  simply  naming  the  letter ;  as, 
angle  A  (Fig.  7). 

But  when  two  or  more  angles  have 
the  same  vertex,  each  angle  is  desig- 
nated by  three  letters,  as  shown  in 
Fig.  8,  and  is  read  by  naming  the 
three  letters,  the  one  at  the  vertex  be- 
tween the  others.  Thus,  the  angle 
DAC  means  the  angle  formed  by  the 
sides  AD  and  AC. 

It  is  often  convenient  to  designate 
an  angle  by  placing  a  small  italic  let- 
ter between  the  sides  and  near  the 
vertex,  as  in  Fig.  9. 


FIG 


42.    Two  angles  are  equal  if  they 
can  be  made  to  coincide. 


FIG.  9. 


43.  If  the  line  AD  (Fig.  8)  is  drawn  so  as  to  divide  the 
angle  BAG  into  two  equal  parts,  BAD  and  CAD,  AD  is 
called  the  bisector  of  the  angle  BAC.  In  general,  a  line  that 
divides  a  geometrical  magnitude  into  two  equal  parts  is  called 
a  bisector  of  it. 


8 


GEOMETRY. 


44.  Two  angles  are  called  ad- 
jacent when  they  have  the  same 
vertex   and  a  common  side  be- 
tween them ;  as,  the  angles  BOD 
and  AOD  (Fig.  10). 

45,  When   one    straight    line 
stands  upon  another  straight  line 
and  makes  the  adjacent  angles 
equal,    each    of  these   angles   is 
called  a  right  angle.     Thus,  the 

«7  J 

equal   angles   DC  A    and   DCB 
(Fig.  11)  are  each  a  right  angle. 


O 

FIG.  10. 


c 

FIQ.  11. 


46,  When  the  sides  of  an  an- 
gle extend  in  opposite  directions, 

so  as  to  be  in  the  same  straight  line,  the  angle  is  called  a 
straight  angle.  Thus,  the  angle  formed  at  C  (Fig.  11)  with 
its  sides  CA  and  CB  extending  in  opposite  directions  from  (7, 
is  a  straight  angle.  Hence  a  right  angle  may  be  defined  as 
half  a  straight  angle. 

47,  A  perpendicular  to  a  straight  line  is  a  straight  line  that 
makes  a  right  angle  with  it.    Thus,  if  the  angle  DC  A  (Fig.  11) 
is  a  right  angle,  DC  is  perpendicular  to  AB,  and  AB  is  per- 
pendicular to  DC. 

48,  The  point  (as  (7,  Fig.  11)  where  a  perpendicular  meets 
another  line  is  called  the  foot  of  the  perpendicular. 


49,    Every  angle  less  than  a  right  an- 
gle is  called  an  acute  angle;  as,  angle  A. 


FIG.  12. 


50.   Every  angle  greater  than  a  right 
angle  and  less  than  a  straight  angle  is  called  an  obtuse  angle; 
as,  angle  C  (Fig.  13). 


DEFINITIONS. 


9 


51i  Every  angle  greater  than  a  straight  angle  and  less 
than  two  straight  angles  is  called  a  reflex  angle;  as,  angle  0 
(Fig.  14). 


FIG.  13. 


FIG.  14. 


52,  Acute,  obtuse,  and  reflex  angles,  in  distinction  from 
right  and  straight  angles,  are  called  oblique  angles  ;  and  inter- 
secting lines  that  are  not  perpendicular  to  each  other  are 
called  oblique  lines. 

53,  When  two  angles  have  the  same  vertex,  and  the  sides 
of  the  one  are  prolongations  of 

the  sides  of  the  other,  they  are 
called  vertical  angles.  Thus,  a 
and  b  (Fig.  15)  are  vertical  an- 
gles. 


FIG.  15. 


54,  Two   angles    are    called 
complementary  when  their  sum 

is  equal  to  a  right  angle ;  and  each  is  called  the  complement 
of  the  other;  as,  angles  DOB  and  DOC  (Fig.  10). 

55,  Two  angles  are  called  supplementary  when  their  sum  is 
equal  to  a  straight  angle ;  and  each  is  called  the  supplement 
of  the  other;  as,  angles  DOB  and  DO  A  (Fig.  10). 


MAGNITUDE  OF  ANGLES. 

56,    The  size  of  an  angle  depends  upon  the  extent  of  opening 
of  its  sides,  and  not  upon  their  length.     Suppose  the  straight 


10 


GEOMETRY. 


line  OC  to  move  in  the  plane  of  the  paper  from  coincidence 

with  OA,  about  the  point  0  as  a  pivot,  to  the  position  0(7; 

then  the  line  OC  describes  or  generates 

the  angle  AOC,  and  the  magnitude  of  the 

angle  AOC  depends  upon  the  amount 

of  rotation  of  the  line  from  the  position 

OA  to  the  position  OC. 

If  the  rotating  line  moves  from  the 
position  OA  to  the  position  OB,  perpen- 
dicular to  OA,  it  generates  the  right 
angle  AOB ;  if  it  moves  to  the  position 
01),  it  generates  the  obtuse  angle  AOD ;  if  it  moves  to  the  posi- 
tion OA1,  it  generates  the  straight  angle  AOA1 ;  if  it  moves  to 
the  position  OB1,  it  generates  the  reflex  angle  AOB1,  indicated 
by  the  dotted  line  ;  and  if  it  continues  its  rotation  to  the  posi- 
tion OA,  whence  it  started,  it  generates  two  straight  angles. 

Hence  the  whole  angular  magnitude  about  a  point  in  a 
plane  is  equal  to  two  straight  angles,  or  four  right  angles;  and 
the  angular  magnitude  about  a  point  on  one  side  of  a  straight 
line  drawn  through  that  point  is  equal  to  one  straight  angle, 
or  two  right  angles. 

Angles  are  magnitudes  that  can  be  added  and  subtracted ; 
they  may  also  be  multiplied  and  divided  by  a  number. 


ANGULAR  UNITS. 

57.  If  we  suppose  00  (Fig.  17)  to 
turn  about  0  from  a  position  coinci- 
dent with  OA  until  it  makes  a  com- 
plete revolution  and  comes  again  into 
coincidence  with  OA,  it  will  describe 
the  whole  angular  magnitude  about 
the  point  0,  while  its  end  point  C 
will  describe  a  curve  called  a  circum- 
ference. 


DEFINITIONS.  11 

58,  By  adopting  a  suitable  unit  of  angles  we  are  able  to 
express  the  magnitudes  of  angles  in  numbers. 

If  we  suppose  00  (Fig.  17)  to  turn  about  0  from  coinci- 
dence with  OA  until  it  makes  one  three  hundred  and  sixtieth 
of  a  revolution,  it  generates  an  angle  at  0,  which  is  taken 
as  the  unit  for  measuring  angles.  This  unit  is  called  a 
degree. 

The  degree  is  subdivided  into  sixty  equal  parts  called 
minutes,  and  the  minute  into  sixty  equal  parts,  called  seconds. 

Degrees,  minutes,  and  seconds  are  denoted  by  symbols. 
Thus,  5  degrees  13  minutes  12  seconds  is  written,  5°  13'  12". 

A  right  angle  is  generated  when  00  has  made  one-fourth 
of  a  revolution  and  is  an  angle  of  90° ;  a  straight  angle  is 
generated  when  OO  has  made  one-half  of  a  revolution  and 
is  an  angle  of  180° ;  and  the  whole  angular  magnitude  about 
0  is  generated  when  00  has  made  a  complete  revolution,  and 
contains  360°. 

The  natural  angular  unit  is  one  complete  revolution.  But 
the  adoption  of  this  unit  would  require  us  to  express  the 
values  of  all  angles  by  fractions.  The  advantage  of  using  the 
degree  as  the  unit  consists  in  its  convenient  size,  and  in  the  fact 
that  360  is  divisible  by  so  many  different  integral  numbers. 


METHOD  OF  SUPERPOSITION. 

59,  The  test  of  the  equality  of  two  geometrical  magnitudes 
is  that  they  coincide  throughout  their  whole  extent. 

Thus,  two  straight  lines  are  equal,  if  they  can  be  so  placed 
that  the  points  at  their  extremities  coincide.  Two  angles  are 
equal,  if  they  can  be  so  placed  that  they  coincide. 

In  applying  this  test  of  equality,  we  assume  that  a  line  may 
be  moved  from  one  place  to  another  without  altering  its  length; 
that  nn  angle  may  be  taken  up,  turned  over,  and  put  down, 
without  altering  the  difference  in  direction  of  its  sides. 


12 


GEOMETRY. 


FIG.  18. 

This  method  enables  us  to  compare  magnitudes  of  the  same 
kind.  Suppose  we  have  two  angles,  ABC  and  DEF.  Let 
the  side  ED  be  placed  on  the  side  BA,  so  that  the  vertex  E 
shall  fall  on  B ;  then,  if  the  side  EF  falls  on  BC,  the  angle 
DEF  equals  the  angle  ABC;  if  the  side  EF  falls  between 
BC  &nd  BA  in  the  direction  BG,  the  angle  DEF  is  less  than 
ABC;  but  if  the  side  EF  falls  in  the  direction  BH,  the  angle 
DEF  is  greater  than  ABC. 

This  method  enables  us  to  add  magnitudes  of  the  same  kind. 

Thus,  if  we  have  two  straight  lines      BC 

AB  and  CD,  by  placing  the  point    ~ ^ 

C  on  B,  and  keeping  CD  in  the   ^ B 

same  direction  with  AB,  we  shall  FIG-  19. 

have  one  continuous  straight  line  AD  equal  to  the  sum  of 

the  lines  AB  and  CD. 

C 

/* 


FIG.  20. 


B 

FIG.  21. 


Again  :  if  we  have  the  angles  ABC  and  DEF,  and  place 
the  vertex  E  on  B  and  the  side  ED  in  the  direction  of  BC,  the 
angle  DEF  will  take  the  position  CBH,  and  the  angles  DEF 
and  ^4.5 (7  will  together  equal  the  angle  ABH. 

If  the  vertex  E  is  placed  on  B,  and  the  side  ED  on  j£Li,  the 
angle  DEF  will  take  the  position  ABF,  and  the  angle  FBC 
will  be  the  difference  between  the  angles  ABC  and  DEF. 


DEFINITIONS. 


13 


FIG.  23. 


SYMMETRY. 

60,  Two  points  are  said  to  be  symmetrical  with  respect  to  a 
third  point,  called  the  centre  of  sym- 

metry,  if  this  third  point  bisects  the    pf —        — h —         — p 
straight  line  which  joins  them.  Thus,  FlQt  22> 

P  and  Pr  are  symmetrical  with  respect  to  0  as  a  centre,  if  G 
bisects  the  straight  line  PP*. 

61.  Two  points  are  said  to  be  sym-  p 
metrical  with  respect  to  a  straight 

line,  called '  the  axis  of  symmetry,  if 

this   straight    line   bisects   at    right        X- 1 -X. 

angles  the  straight  line  which  joins 
them.  Thus,  P  and  P'  are  symmet- 
rical with  respect  to  XX1  as  an  axis, 
if  XX'  bisects  PP'  at  right  angles. 

62.  Two  figures  are  said  to  be  sym- 
metrical with  respect  to  a  centre  or 
an  axis  if  every  point  of  one  has  a 
corresponding  symmetrical  point  in 
the  other.     Thus,  if  every  point  in 
the  figure  A'J3'C'  has  a  symmetrical 
point  in  ABC,  with  respect  to  D  as 
a  centre,  the  figure  A'H'C'  is   sym- 
metrical to  ABO  with  respect  to  D 
as  a  centre. 

63,  If  every  point  in   the  figure 
A'B'C1  has  a  symmetrical  point  in 
ABO,  with  respect  to   XX*  as  an 
axis,  the  figure  A'B'C1  is  symmetri- 
cal to  ABC  with  respect  to  XX1  as 
an  axis. 


14 


GEOMETRY. 


64,  A  figure  is  symmetrical  with  re- 
spect to  a  point,  if  the  point    bisects 
every   straight   line   drawn   through  it 
and  terminated  by  the  boundary  of  the 
figure. 

65,  A  plane  figure  is  symmetrical  with 
respect  to  a  straight  line,  if  the   line 
divides  it  into  two  parts,  which  are  sym- 
metrical with   respect  to   this  straight 
line. 

MATHEMATICAL  TERMS. 


FIG.  27. 


66,  A  proof  or  demonstration  is  a  course  of  reasoning  by 
which   the   truth   or   falsity   of  any   statement  is    logically 
established. 

67,  A  theorem  is  a  statement  to  be  proved. 

68,  A  theorem  consists  of  two  parts :   the  hypothesis,  or 
that  which  is  assumed ;  and  the  conclusion,  or  that  which  is 
asserted  to  follow  from  the  hypothesis. 

69,  An  axiom  is  a  statement  the  truth  of  which  is  admitted 
without  proof. 

70,  A  construction  is  a  graphical  representation  of  a  geo- 
metrical figure. 

71,  A  problem  is  a  question  to  be  solved. 

72,  The  solution  of  a  problem  consists  of  four  parts  : 

(1)  The  analysis,  or  course  of  thought  by  which  the  con- 
struction of  the  required  figure  is  discovered ; 

(2)  The  construction  of  the  figure  with  the  aid  of  ruler  and 
compasses ; 

(3)  The  proof  that  the  figure  satisfies  all  the  given  condi- 
tions; 


DEFINITIONS.  15 

(4)    The  discussion  of  the  limitations,   which  often  exist, 
within  which  the  solution  is  possible. 

73,  A  postulate  is  a  construction  admitted  to  be  possible. 

74,  A  proposition  is  a  general  term  for  either  a  theorem  or 
a  problem. 

75,  A  corollary  is  a  truth  easily  deduced  from  the  propo- 
sition to  which  it  is  attached. 

76,  A  scholium  is  a  remark  upon  some  particular  feature 
of  a  proposition. 

77,  The  converse  of  a  theorem  is  formed  by  interchanging 
its  hypothesis  and  conclusion.      Thus, 

If  A  is  equal  to  B,  O  is  equal  to  D.    (Direct.) 
If  C  is  equal  to  D,  A  is  equal  to  B.    (Converse.) 

78,  The  opposite  of  a  proposition  is  formed  by  stating  the 
negative  of  its  hypothesis  and  its  conclusion.     Thus, 

If  A  is  equal  to  B,  C  is  equal  to  D.    (Direct.) 

If  A  is  not  equal  to  B,  C  is  not  equal  to  D.    (Opposite.) 

79,  The  converse  of  a  truth  is  not  necessarily  true.     Thus, 
Every  horse  is  a  quadruped  is  a  true  proposition,  but  the 
converse,  Every  quadruped  is  a  horse,  is  not  true. 

80,  If  a  direct  proposition  and  its  converse  are  true,   the 
opposite  proposition  is  true ;  and  if  a  direct  proposition  and  its 
opposite  are  true,  the  converse  proposition  is  true. 

81,  POSTULATES. 
Let  it  be  granted  — 

1.  That  a  straight  line  can  be  drawn  from  any  one  point 
to  any  other  point. 

2.  That  a  straight  line  can  be  produced  to  any  distance, 
or  can  be  terminated  at  any  point. 

3.  That  a  circumference  may  be  described  about  any  point 
as  a  centre  with  a  radius  of  given  length, 


16  GEOMETRY. 

82,  AXIOMS. 

1.  Things  which  are  equal  to  the  same  thing  are  equal  to 
each  other. 

2.  If  equals  are  added  to  equals  the  sums  are  equal. 

3.  If  equals  are  taken  from  equals  the  remainders  are  equal. 

4.  If  equals  are  added  to  unequals  the  sums  are  unequal, 
and  the  greater  sum  is  obtained  from  the  greater  magnitude. 

5.  If  equals  are  taken  from  unequals  the  remainders  are 
unequal,    and   the  greater   remainder  is  obtained  from  the 
greater  magnitude. 

6.  Things  which  are    double  the  same   thing,    or   equal 
things,   are  equal  to  each  other. 

7.  Things  which  are  halves  of  the  same  thing,  or  of  equal 
things,  are  equal  to  each  other. 

8.  The  whole  is  greater  than  any  of  its  parts. 

9.  The  whole  is  equal  to  all  its  parts  taken  together. 

83,  SYMBOLS  AND  ABBREVIATIONS. 

4-  increased  by.  0  circle.     ®  circles. 

—  diminished  by.  Def.  .  .  .  definition. 

X  multiplied  by.  Ax.    .  .  .  axiom. 

-T-  divided  by.  Hyp-    .  .  hypothesis. 

=  is  (or  are)  equal  to.  Cor.  .  .  .  corollary. 

=z=  is  (or  are)  equivalent  to.  Adj.  .  .  .  adjacent. 

>  is  (or  are)  greater  than.  Iden.    .  .  identical. 

<  is  (or  are)  less  than.  Cons.    .  .  construction. 

.-.  therefore.  Sup.  .  .  .  supplementary. 

/_  angle.  Sup. -adj.  supplementary. 

^1  angles.  Ext. -int.  exterior-interior. 

_L  perpendicular.  Alt.-int.    alternate-interior. 

Jl  perpendiculars.  Ex.    .  .  .  exercise. 

II   parallel.  rt right. 

US  parallels.  st straight. 

A  triangle.  Q.E.D.  .  .  quod  erat  demonstrandum, 

&  triangles.  which  was  to  be  proved. 

n  parallelogram.  Q.E.F.  .  .  quod   erat  faciendum, 
/T7  parallelograms.  which  was  to  be  done. 


PLANE  GEOMETRY. 

BOOK   I. 
THE    STRAIGHT    LI1TE. 


PROPOSITION  I.     THEOREM. 
84,  All  straight  angles  are  equal. 


D  -  -  r 

Let  Z.BCA  and  Z  FED  be  any  two  straight  angles. 
To  prove  AECA^Z.  FED. 

Proof,   Apply  the  /.BCAio  the  Z  FED,  so  that  the  vertex 
C  shall  fall  on  the  vertex  E,  and  the  side  CB  on  the  side  EF. 

Then  CA  will  coincide  with  ED, 
(because  EC  A  and  FED  are  straight  lines  and  have  two  points  common). 

Therefore  the  Z  BCA  is  equal  to  the  Z  FED.  §  59 

Q.  E.  D. 

85,  COR.  1.   All  right  angles  are  equal.  Ax.  7. 

86,  COR.  2.    The  angular  units  have  constant  values. 

87,  COR.  3.    The  complements  of  equal  angles  are  equal.  Ax.  3. 

88,  COR.  4.    The  supplements  of  equal  angles  are  equal.  Ax.  3. 

89,  COR.  5.  At  a  given  point  in  a  given  straight  line  one 
perpendicular  •,  and  only  one,  can  be  erected. 

HINT.    Consider  the  given  point  as  the  vertex  of  a  straight  angle,  and 
draw  the  bisector  of  the  angle. 


18  PLANE   GEOMETKY. —  BOOK   I. 


PROPOSITION  II.    THEOREM. 

90.  If  two  adjacent  angles  have  their  exterior  sides 
in  a  straight  line,  these  angles  are  supplements  of 
each  other. 


Let  the  exterior  sides  OA  and  OB  of  the  adjacent 
A  AOD  and  BOD  be  in  the  straight  line  AB. 

To  prove  A  AOD  and  BOD  supplementary. 

Proof.                    AOE  is  a  straight  line.  Hyp. 

/.  the  Z  AOB  is  a  st.  Z.  §  46 

But         the  Z  AOD  +  Z  BOD  =  the  st.  Z  AOB.  Ax.  9 

/.  the  A  AOD  and  BOD  are  supplementary.  §  55 

Q.  E.  D. 

91.  SCHOLIUM.   Adjacent  angles  that  are  supplements  of 
each  other  are  called  supplementary -adjacent  angles. 

92.  COR.    Since  the  angular  magnitude   about  a  point  is 
neither  increased  nor  diminished  by  the  number  of  lines  which 
radiate  from  the  point,  it  follows  that, 

The  sum  of  all  the  angles  about  a  point  in  a  plane  is  equal 
to  two  straight  angles,  or  four  right  angles. 

The  sum  of  all  the  angles  about  a  point  on  the  same  side  of  a 
straight  line  passing  through  the  point  is  equal  to  a  straight 
angle,  or  two  right  angles. 


THE   STRAIGHT   LINE.  19 


PROPOSITION  III.    THEOREM. 

93.  CONVERSELY  :  If  two  adjacent  angles  are  suppl&* 
merits  of  each  other,  their  exterior  sides  lie  in  the 
same  straight  line. 


AC 
Let  the  adjacent  A  OCA 

To  prove  A  C  and  CB  in  the  same  straight  line. 

Proof,   Suppose  £Fto  be  in  the  same  line  with  -4  C.       §  81 

Then  Z  OCA  +  Z  OCF=  2  rt.  A,  §  90 

(being  sup.-adj.  A). 

But  Z  OCA  +  Z  OCB  =  2  rt.  A.  Hyp. 

.-.  Z  OCA  +  Z  OCF=  Z  OCA  +  Z  OCB.       Ax.  1 
Take  away  from  each  of  these  equals  the  common  Z  OCA. 
Then  Z  OCF=  Z  OCB.  Ax.  3 

.'.  CB  and  CF  coincide. 
.'.  ^C'and  CB  are  in  the  same  straight  line.       Q.E.D. 

94,  SCHOLIUM.  Since  Propositions  II.  and  III.  are  true, 
their  opposites  are  true  ;  namely,  §  80 

If  the  exterior  sides  of  two  adjacent  angles  are  not  in  a 
straight  line,  these  angles  are  not  supplements  of  each  other. 

If  two  adjacent  angles  are  not  supplements  of  each  other, 
thevr  exterior  sides  are  not  in  the  same  straight  line. 


20  PLANE   GEOMETRY. — BOOK   L 


PROPOSITION  IV,     THEOREM. 

95,  If  one  straight  line  intersects  another  straight 
tine,  the  vertical  angles  are  equal. 


Let  line  OP  cut  AB  at  C. 
To  prove  Z  0GB  =  Z  AGP. 

Proof,  Z  OCA  +  Z  OCB  =  2  rt,  A,  §  90 

(being  sup.  -adj.  A). 

Z  OCA  +  Z  AGP=  2  rt.  A,  §  90 

(being  sup.  -adj.  /«). 

.'.  Z  OCA  +  Z  005  =  Z  004  +  Z  ACP.  Ax.  1 
Take  away  from  each  of  these  equals  the  common  Z  OO4. 
Then  Z  005  =  Z  ACP.  Ax.  3 

In  like  manner  we  may  prove 

Q.E.D. 


96,  COR.  If  one  of  the  four  angles  formed  by  the  intersection 
of  two  straight  lines  is  a  right  angle,  the  other  three  angles  are 
right  angles. 


THE   STKAIGHT   LINE.  21 


PROPOSITION  V.     THEOREM. 

97.   From  a  point  without  a  straight  line  one  per- 
pendicular, and  only  one,  can  be  drawn  to  this  line. 

P 

A 


D\ 


P' 

Let  P  be  the  point  and  AB  the  line. 

To  prove  that  one  perpendicular,  and  only  one,  can  be  drawn 
from  P  to  AB. 

Proof,  Turn  the  part  of  the  plane  above  AB  about  AB  as 
an  axis  until  it  falls  upon  the  part  below  AB,  and  denote  by 
P1  the  position  that  P  takes. 

Turn  the  revolved  plane  about  AB  to  its  original  position, 
and  draw  the  straight  line  PP1,  cutting  AB  at  C. 

Take  any  other  point  D  in  AB,  and  draw  PD  and  P'D. 

Since  POP'  is  a  straight  line,  PDF  is  not  a  straight  line. 
(Between  two  points  only  one  straight  line  can  be  drawn.) 

.'.  Z  PCT'is  a  st.  Z,  and  Z  PDP  is  not  a  st.  Z. 
Turn  the  figure  PCD  about  AB  until  P  falls  upon  P. 
Then  OP  will  coincide  with  CP,  and  DP  with  DP. 
.'.  Z  PCD  =  Z  PCD,  and  Z  PDC=  Z  PDC.  §  59 

.'.  Z  PCD,  the  half  of  st.  Z  POP',  is  a  rt.  Z  ;  and  Z  PDC, 
the  half  of  Z  PDP,  is  not  a  rt.  Z. 

.-.  P(7is  _L  to  AB,  and  PD  is  not  _L  to  ^ A  §  47 

.'.  one  JL,  and  only  one,  can  be  drawn  from  P  to  AB. 

Q.E.  D. 


22  PLANE    GEOMETRY.  —  BOOK!    I. 

PARALLEL  LINES. 

98,  DEF.    Parallel  lines  are   lines  which   lie  in  the  same 
plane  and  do  not  meet  however  far  they  are  prolonged  in  both 
directions. 

99,  Parallel  lines  are  said  to  lie  in  the  same  direction  when 
they  are  on  the  same  side  of  the  straight  line  joining  their  ori- 
gins, and  in  opposite  directions  when  they  are  on  opposite  sides 
of  the  straight  line  joining  their  origins. 

PROPOSITION  VI. 

100,  Two  straight  lines  in  the  same  plane  perpen- 
dicular to  the  same  straight  line  are  parallel. 


-B 


Let  AB  and  CD  be  perpendicular  to  AC. 

To  prove  AB  and  CD  parallel. 

Proof,  If  AB  and  CD  are  not  parallel,  they  will  meet  if 
sufficiently  prolonged,  and  we  shall  have  two  perpendicular 
lines  from  their  point  of  meeting  to  the  same  straight  line ; 
but  this  is  impossible.  §  97 

(From  a  given  point  without  a  straight  line,  one  perpendicular,  and  only 
one,  can  be  drawn  to  the  straight  line.) 

.'.  AB  and  CD  are  parallel.  Q.E.D. 

REMARK.  Here  the  supposition  that  AB  and  CD  are  not  parallel  leads 
to  the  conclusion  that  two  perpendiculars  can  be  drawn  from  a  given 
point  to  a  straight  line.  The  conclusion  is  false,  therefore  the  supposi- 
tion is  false;  but  if  it  is  false  that  AB  and  CD  are  not  parallel,  it  is  true 
that  they  are  parallel.  This  method  of  proof  is  called  the  indirect 
method. 

101,  COR.  Through  a  given  point,  one  straight  line,  and  only 
one,  can  be  drawn  parallel  to  a  given  straight  line. 


PARALLEL   LINES. 


23 


PROPOSITION  VII.     THEOREM. 

102,  If  a  straight  line  is  perpendicular  to  one  of 
two  parallel  lines,  it  is  perpendicular  to  the  other. 


H 

o 


E 


G 


F 


K 


Let  AB  and  EF  be  two  parallel  lines,  and  let  HK  be 
perpendicular  to  AB,  and  cut  EF  at  C. 

To  prove  HK\_EF. 

Proof,    Suppose  MN  drawn  through  (7J_  to  HK. 

Then  MNis  \\  to  AB,  §  100 

(two  lines  in  the  same  plane  _L  to  a  given  line  are  parallel). 
But  JSFis\\  to  AB.  Hyp. 

.-.  EF  coincides  with  MN,  §  101 

(through  the  same  point  only  one  line  can  be  drawn  \\  to  a  given  line). 

.*.  EFis  _L  to  HK, 
that  is,  HK\s  J_  to  EF.  Q.E.D. 

103,    If  two  straight  lines  AB 
and  CD  are  cut  by  a  third  line      . 
EF,    called    a    transversal,    the 
eight  angles  formed  are  named 
as  follows : 

The  angles  a,  d,  f,  g  are  called 
interior ;  b,  c,  e,  h  are  called  ex- 
terior angles. 

The  angles  d  and/,  or  a  and  g,  are  called  alt. -int.  angles. 

The  angles  b  and  A,  or  c  and  e,  are  called  alt.-ext.  angles. 

The  angles  b  and  /,  c  and  g,  a  and  e,  or  d  and  A,  are  called 
ext.-int.  angles. 


24  PLANE   GEOMETRY.  —  BOOK   I. 


PROPOSITION  VIII.    THEOREM. 

104,  If  two  parallel  straight  lines  are  cut  by  a  third 
straight  line,  the  alternate-interior  angles  are  equal. 


E 


/5_  /  _  1      __  __  TT 

~" 


Let  EF  and  GH  be  two  parallel  straight  lines  cut  by 
the  line  BO. 

To  prove  Z  B  =  Z  0. 

Proof.  Through  0,  the  middle  point  of  EC,  suppose  AD 
drawn  J_  to  GH. 

Then  AD  is  likewise  1  to  EF,  §  102 

(a  straight  line  _L  to  one  of  two  Us  is  JL  to  the  other), 

that  is,  CD  and  BA  are  both  J.  to  AD. 

Apply  figure  COD  to  figure  BOA,  so  that  OD  shall  fall 
on  OA. 

Then  OC  will  fall  on  OB,  §  95 

(since  Z.  COD  =  Z.  BOA,  being  vertical  A)  ; 

and  the  point  C  will  fall  upon  B, 

(since  0(7  =  OB  by  construction). 

Then      the  J.  OD  will  coincide  with  the  ±.BA,  §  97 

(from  a  pvlnt  without  a  straight  line  only  one  JL  to  that  line  can  be  drawn]. 

.*.  Z  OCD  coincides  with  Z  OB  A,  and  is  equal  to  it.  §59 

_  Q.  E.  D. 

Ex.  1.  Find  the  value  of  an  angle  if  it  is  double  its  complement  ;  if 
it  is  one-fourth  of  its  complement. 

Ex.  2.  Find  the  value  of  an  angle  if  it  is  double  its  supplement  ;  if  it 
is  one-third  of  its  supplement. 


PARALLEL   LINES./  25 


PROPOSITION  IX.    THEOREM. 

105,  CONVERSELY:  When  two  straight  lines  are  cut 
by  a  third  straight  line,  if  the  alternate-interior  an- 
gles are  equal,  the  two  straight  lines  are  parallel. 


M           A 

7 

B 

N 

/ 

n 

n 

Let  EF  cut  the  straight  lines  AB  and  CD  in  the  points 
H  and  K,  and  let  the  Z.AHK  =  £HKD. 

To  prove  AB  II  to  CD. 

Proof,     Suppose  JOT  drawn  through  H II  to  CD\         §  101 

then  Z  MHK=  Z  HKD,  §  104 

(being  alt. -int.  A  of  II  lines). 

But  Z  AHK=  Z  HKD.  Hyp. 

/.  Z  MHK=  Z  AHK.  Ax.  1 

.*.  the  lines  JOT  and  AB  coincide. 

But  MN\R  II  to  CD.  Cons. 

/.  AB,  which  coincides  with  MN,  is  II  to  CD. 

Q.E.  D. 


Ex.  3.   How  many  degrees  in  the  angle  formed  by  the  hands  of  a 
clock  at  2  o'clock ?  3  o'clock?  4  o'clock?  6  o'clock? 


26  PLANE   GEOMETEY.  —  BOOK    I. 


PROPOSITION  X.     THEOREM. 

106,  If  two  parallel  lines  are  entity  a  third  straight 
line,  the  exterior-interior  angles  are  equal. 


-D 


Let  AB  and  CD  "be  two  parallel  lines  cut  by  the 
straight  line  EF,  in  the  points  H  and  K. 

To  prove  Z  EHB  =  Z  HKD. 

Proof,  Z  EHB  =  Z  AHK,  §  95 

(being  vertical  A). 

•  But  Z  AHK=  Z  HKD,  §  104 

(being  alt.-int.  Aofll  lines). 

.\  Z.  EHB  = /.  HKD.  Ax.  1 

In  like  manner  we  may  prove 

Z  EH  A  =  Z  HKO. 

Q.  E.  D. 

107,   COR.   The  alternate-exterior  angles  EHB  and  CKF, 
and  also  A  HE  and  DKF,  are  equal. 


Ex.  4.  If  an  angle  is  bisected,  and  if  a  line  is  drawn  through  the 
vertex  perpendicular  to  the  bisector,  this  line  forms  equal  angles  with 
the  sides  of  the  given  angle. 

Ex.  5.  If  the  bisectors  of  two  adjacent  angles  are  perpendicular  to 
each  other,  the  adjacent  angles  are  supplementary. 


PARALLEL   LINES.  27 


PROPOSITION  XI.    THEOREM. 

108,  CONVERSELY  :  When  two  straight  lines  are  cut 
by  a  third  straight  line,  if  the  exterior-interior  an- 
gles are  equal,  these  two  straight  lines  are  parallel. 


M        A  /  B  N 

C 


Let  EF  cut  the  straight  lines  AB  and  CD  in  the  points 
H  and  K,  and  let  the  ^.EHB  =  ^.HKD. 

To  prove  AB  II  to  CD. 

Proof,     Suppose  MN  drawn  through  H II  to  CD.         §  101 

Then  Z.  EHN=  Z.  HKD,  §  106 

(being  ext.-int.  A  of  II  lines). 

But  Z  EHB  =  Z  HKD.  Hyp. 

.:Z.EHB  =  £EHN.  Ax.  1 

.'.the  lines  _$/7Vand  AB  coincide. 

But  MNis  II  to  CD.  Cons. 

.*.  AB,  which  coincides  with  MN,  is  II  to  CD. 

Q.  E.  D. 


Ex.  6.  The  bisector  of  one  of  two  vertical  angles  bisects  the  other. 
Ex.  7.   The  bisectors  of  the  two  pairs  of  vertical  angles  formed  by 
two  intersecting  lines  are  perpendicular  to  each  other. 


28  PLANE   GEOMETRY. — BOOK   I. 


PROPOSITION  XII.    THEOREM. 

109.  If  two  parallel  lines  are  cut  by  a  third  straight 
line,  the  sum  of  the  two  interior  angles  on  the  same 
side  of  the  transversal  is  equal  to  two  ri^M  angles. 


Let  AB  and  CD  be   two  parallel  lines  cut  by  the 
straight  line  EF  in  the  points  H  and  K. 

To  prove         Z  BHK+  Z  HKD  =  2  rt.  A 

Proof.  Z  EHB  +  Z  EHK  =  2  rt.  A,  §  90 

^  (being  sup.-adj.  A). 

But  Z  EHB  =  Z  HKD,  §  106 

(being  ext.-int.  A  o/\\  lines). 

Substitute  Z  HKD  for  Z  EHB  in  the  first  equality ; 
then  Z  BHK+  Z  HKD  =  2  rt.  A. 

Q.  E.  D. 


Ex.  8.   If  the  angle  AHE  is  an  angle  of  135°,  find  the  number  of 
degrees  in  each  of  the  other  angles  formed  at  the  points  JJ  and  K. 

Ex.  9.    Find  the  angle  between  the  bisectors  of  adjacent  complemen- 
tary angles. 


PARALLEL  LINES.  29 


PROPOSITION  XIII.     THEOREM. 

110,  CONVERSELY  :  When  two  straight  lines  are  cut 
by  a  third  straight  line,  if  the  two  interior  angles  on 
the  same  side  of  the  transversal  are  together  equal  to 
two  right  angles,  then  the  two  straight  lines  are 
parallel. 


MA  /  B  N 


-D 


F 

Let  EF  cut  the  straight  lines  AB  and  CD  in  the  points 
H  and  K,  and  let  the  Z.BHK  +  /LUKD  equal  two  right 
angles. 

To  prove  AB  II  to  CD. 

Proof,     Suppose  MN  drawn  through  H  II  to  CD. 

Then  Z  NHK+  Z  HKD  =  2  rt.  Zs,  §  109 

(being  two  interior  Aofllson  the  same  side  of  the  transversal). 

But  Z  B HK+  Z  HKD  =  2  rt.  Zs.  Hyp. 

.\Z.NHK+^HKD  =  /.BHK+/.HKD.    Ax.  1 

Take  away  from  each  of  these  equals  the  common  Z  HKD ; 

then  Z  NHK=  Z  BHK.  Ax.  3 

.'.  the  lines  AB  and  MN  coincide. 

But  MNi*  II  to  CD.  Cons. 

/.  AB,  which  coincides  with  MN,  is  II  to  CD. 

Q.E.  D. 


30  PLANE  GEOMETRY.  —  BOOK  I. 


PROPOSITION  XIV.     THEOREM. 

Ill,  Two  straight  lines  which  are  parallel  to  a  third 
straight  line  are  parallel  to  each  other. 


Let  AB  and  CD  fee  parallel  to  EF. 

To  prove  AB  II  to  CD. 

Proof,  Suppose  HK  drawn  J_  to  EF.  §  97 

Since  CD  and  EF  are  II,  UK  is  J.  to  CD,        §  102 

(if  a  straight  line  is  JL  to  one  of  two  Us,  it  is  _L  £0  the  other  also). 

Since  ^^  and  EF  are  II,  JZfiTis  also  _L  to  ^£.    §  102 


(each  being  a  rt.  /). 
.'.  AB  is  II  to  CD,  §  108 

(when  two  straight  lines  are  cut  by  a  third  straight  line,  if  the  ext.-int.  A 
are  equal,  the  two  lines  are  parallel). 

Q.  E.  o. 

Ex.  10.  It  has  been  shown  that  if  two  parallels  are  cut  by  a  trans* 
versal,  the  alternate-interior  angles  are  equal,  the  exterior-interior  angles 
are  equal,  the  two  interior  angles  on  the  same  side  of  the  transversal  are 
supplementary.  State  the  opposite  theorems.  State  the  converse  theo- 


PARALLEL   LINES.  31 


PKOPOSITION'  XV.     THEOREM. 

112,  Two  angles  whose  sides  are  parallel,  each  to 
each,  are  either  equal  or  supplementary. 


F 
Let  AB  be  parallel  to  EF,  and  BC  to  MN. 

To  prove  Z  ABO  equal  to  Z  EHN,  and  to  Z  MHF,  and 
supplementary  to  Z  EHM  and  to  Z  NHF. 

Proof,  Prolong  (if  necessary)  BC  and  FE  until  they  inter- 
sect at  D.  §  81  (2) 

Then  Z.B  =  /.EDC,  §106 

and  Z  DHN=  Z  EDO.  §  106 

(being  ext.-int.  A  of  II  lines), 

.•.£B  =  Z.DHN;  Ax.  1 

and  Z  B  =  Z  MHF  (the  vert.  Z  of  DUN}. 

Now  Z  DHNis  the  supplement  of  Z  EHM  and  Z 

.'.ZJB,  which  is  equal  to  Z 
is  the  supplement  of  Z  EHM  and  of  Z 

Q.  E.  D. 

REMARK.  The  angles  are  equal  when  both  pairs  of  parallel  sides 
extend  in  the  same  direction,  or  in  opposite  directions,  from  their  ver- 
tices ;  the  angles  are  supplementary  when  two  of  the  parallel  sides  extend 
in  the  same  direction,  and  the  other  two  in  opposite  directions,  from  their 
vertices. 


32 


PLANE   GEOMETEY.  —  BOOK   I. 


PROPOSITION  XVI.    THEOREM. 

113.  Two  angles  whose  sides  are  perpendicular,  each 
to  each,  are  either  equal  or  supplementary. 

G 

Tf 

D\ 


K 


\ 


F 


Let  AB  "be  perpendicular  to  FD,  and  AC  to  GL 

To  prove  Z  BAC  equal  to  Z  DFG,  and  supplementary  to 


Proof,   Suppose  AK  drawn  _L  to  AB,  and  AH  _L  to  AQ. 

Then  AKis\\toFD,audAHtoIG,  §100 

(two  lines  _L  to  the  same  line  are  parallel). 

.\Z.DFG  =  ^KAH,  §112 

(two  angles  are  equal.  whose  sides  are  I!  and  extend  in  the  same  direction 
from  their  vertices). 

The  Z  BAK  is  a  right  angle  by  construction. 

.'.  Z  BAH  is  the  complement  of  Z  KAH. 
The  Z  CAHis  a  right  angle  by  construction. 

/.  Z  BAH  is  the  complement  of  Z  BAG. 

§87 


(complements  of  equal  angles  are  equal). 
.'.ZDFG--=ZBAC. 


Ax.  1 


.  DPI,  the  supplement  of  Z  DFG,  is  also  the  supplement 


Q.E.D. 


REMARK.   The  angles  are  equal  if  both  are  acute  or  both  obtuse  ;  they 
are  supplementary  if  one  is  acute  and  the  other  obtuse. 


PERPENDICULAR  AND  OBLIQUE   LINES.  33 

PERPENDICULAR  AND  OBLIQUE  LINES. 

PROPOSITION  XVII.     THEOREM. 

114,    The  perpendicular  is  the  shortest  Line  that  can 
be  drawn  from  a  point  to  a  straight  Line. 


Let  AB  "be  the  given  straight  line,  P  the  given  point, 
PC  the  perpendicular,  and  PD  any  other  line  drawn 
from  P  to  AB. 

To  prove  PC  <  PD. 

Proof,  Produce  PC  to  P',  making  OP'=  PC]  and  draw  DP1. 
On  AB  as  an  axis,  fold  over  CPD  until  it  conies  into  the 
plane  of  CP'D. 

The  line  OP  will  take  the  direction  of  CP', 

(since  Z  PCD  =  Z  P'CZ),  each  being  a  rt.  Z  by  hyp.). 

The  point  P  will  fall  upon  the  point  Pr, 

(since  PC=  PC  by  cons.). 
/.line  PD  =  line  P'D, 


and  PC  +  CP'  =2  PC.  Cons. 

But  PC  +  CP1  <PD  +  DP\ 

(a  straight  line  is  the  shortest  distance  between  two  points). 

.\2PC<2PDt  or  PC<PD.  Q.E.O. 


34  PLANE  GEOMETRY.  —  BOOK  X, 

115,  SCHOLIUM.  The  distance  of  a  point  from  a  line  is  under- 
stood to  mean  the  length  of  the  perpendicular  from  the  point 
to  the  line. 

PROPOSITION  XVIII.     THEOREM. 

116,  Two  oblique  lines  drawn  from  a  point  in  a 
perpendicular  to  a  given  line,  cutting  off  equal  dis- 
tances from  the  foot  of  the  perpendicular f  are  equal. 


A  F  0 

Let  FC  be  the  perpendicular,  and  CA  and  CO  two 
oblique  lines  cutting  off  equal  distances  from  F. 

To  prove  OA  =  CO. 

Proof,  Fold  over  CFA,  on  ClFas  an  axis,  until  it  comes  into 
the  plane  of  CFO. 

FA  will  take  the  direction  of  FO, 
(since  Z.  OF  A  =  /.  CFO,  each  being  a  rt.  Z.  by  hyp.). 

Point  A  will  fall  upon  point  0, 
(since  FA  =  FO  by  hyp.). 

/.line  G4  =  line  CO, 
(their  extremities  being  the  same  points).  Q.  E.  D. 

117,  COR.  Two  oblique  lines  drawn  from  a  point  in  a  per- 
pendicular to  a  given  line,  cutting  off  equal  distances  from  the 
foot  of  the  perpendicular,  make  equal  angles  with  the  given  line, 
and  also  with  the  perpendicular. 


PERPENDICULAE  AND   OBLIQUE   LINES.  35 


PROPOSITION  XIX.    THEOREM. 

118.  The  sum  of  two  lines  drawn  from  a  point  to 
the  extremities  of  a  straight  line  is  greater  than  the 
sum  of  two  other  lines  similarly  drawn,  but  included 
ly  them. 

C 


Let  CA  and  GB  be  two  lines  drawn  from  the  point  C 
to  the  extremities  of  the  straight  line  AB.  Let  OA  and 
OB  be  two  lines  similarly  drawn,  but  included  by  CA 
and  CB. 

To  prove  CA+CB>OA  +  OB. 

Proof.      Produce  AO  to  meet  the  line  CB  at  K 

Then  AC+CE>OA  +  OE, 

(a  straight  line  is  the  shortest  distance  between  two  points), 

and  BE+OE>BO. 

Add  these  inequalities,  and  we  have 
CA  +  CE+  BE+  OE>  OA  +  OE+  OB. 

Substitute  for  CE+  BEiis  equal  CB, 
and  take  away  OE  from  each  side  of  the  inequality. 
We  have  CA+CB>OA+OB.  Ax.  5 


36 


PLANE   GEOMETRY.  —  BOOK   I. 


PROPOSITION  XX.     THEOREM. 

119.  Of  two  oblique  lines  drawn  from  the  same 
point  in  a  perpendicular,  cutting  off  unequal  dis- 
tances from  the  foot  of  the  perpendicular f  the  more 
remote  is  the  greater. 

o 


F\ 


C     G 


-B 


D 

Let  OC  be  perpendicular  to  AB,  OG  and  OE  two  oblique 
lines  to  AB,  and   CE  greater  than  CG. 

To  prove  OE  >  OG. 

Proof.        Take  OF  equal  to  CG,  and  draw  OF. 

Then  OF=  OG,  §  116 

(two  oblique  lines  drawn  from  a  point  in  a  J_,  cutting  off  equal  distances 
from  the  foot  of  the  J_,  are  equal). 

Prolong  OC  to  Z>,  making  CD  =00. 

Draw  ED  and  FD. 

Since  Alt  is  J_  to  OD  at  its  middle  point, 
FO  =  FD,  and  EO  =  ED, 
But  OE  +  ED  >  OF+  FD, 


§116 
§118 


(the  sum  of  two  oblique  lines  drawn  from  a  point  to  the  extremities  of  a 
straight  line  is  greater  than  the  sum  of  two  other  lines  similarly  drawn, 
but  included  by  them). 

.'.  20E>  20F,  or  OE >  OF. 


But  OF=  OG.     Hence  OE  >  OG. 


Q.  E.  D. 


120,  COR.  Only  two  equal  straight  lines  can  be  drawn  from 
a  point  to  a  straight  line ;  and  of  two  unequal  lines,  the  greater 
cuts  off  the  greater  distance  from  the  foot  of  the  perpendicular. 


PERPENDICULAR   AND   OBLIQUE   LINES.  37 


PROPOSITION  XXI.    THEOREM. 

121.  Two  equal  oblique  lines,  drawn  from  the  same 
point  in  a  perpendicular,  cut  off  equal  distances  from 
the  foot  of  the  perpendicular. 

c 


E  F  K 

Let  CF  "be  the  perpendicular,  and  CE  and  OK  be  two 
equal  oblique  lines  drawn  from  the  point  C  to  AB. 

To  prove  FE=FK. 

Proof,   Fold  over  CFA  on  CF  as  an  axis,  until  it  comes  into 
the  plane  of  CFB. 

The  line  FE  will  take  the  direction  FK, 
(since  Z  CFE=Z.  CFK,  each  being  a  rt.  Z.  by  hyp.). 

Then  the  point  E  must  fall  upon  the  point  Kt 


Otherwise  one  of  these  oblique  lines  must  be  more  remote 
from  the  perpendicular,  and  therefore  greater  than  the  other  ; 
which  is  contrary  to  the  hypothesis  that  they  are  equal.  §  119 

Q.E.  D. 

Ex.  11.  Show  that  the  bisectors  of  two  supplementary-  adjacent 
angles  are  perpendicular  to  each  other. 

Ex.  12.  Show  that  the  bisectors  of  two  vertical  angles  form  one 
straight  line. 

Ex.  13.  Find  the  complement  of  an  angle  containing  26°  52'  37". 
Find  the  supplement  of  the  same  angle. 


38 


PLANE   GEOMETRY. — BOOK  I. 


PROPOSITION  XXII.     THEOREM. 

122,  Every  point  in  the  perpendicular,  erected  at 
the  middle  of  a  given  straight  line,  is  equidistant 
from  the  extremities  of  the  line,  and  every  point  not 
in  the  perpendicular  is  unequally  distant  from  the 
extremities  of  the  line. 


Let  PR  be  a  perpendicular  erected  at  the  middle  of 
the  straight  line  AB,  0  any  point  in  PR,  and  G  any 
point  without  PR. 

Draw  OA  and  OB,  CA  and  CB. 
To  prove  OA  and  OB  equal,  CA  and  CB  unequal. 
Proof.  PA  =  PB.  Hyp. 

§116 


(two  oblique  lines  drawn  from  the  same  point  in  a  _L,  cutting  off  equal  dis 
tances  from  the  foot  of  the  _L,  are  equal). 

Since  C  is  without  the  perpendicular,  one  of  the  lines,  CA 
or  CB,  will  cut  the  perpendicular. 

Let  CA  cut  the  J_  at  D,  and  draw  DB. 

Then  DB  =  DA,  §  116 

(two  oblique  lines  drawn  from  the  same  point  in  a  J_,  cutting  off  equal  dis- 
tances from  the  foot  of  the  _L,  are  equal). 

But  CB<CD+DB, 

(a  straight  line  is  the  shortest  distance  between  two  points). 
Substitute  in  this  inequality  DA  for  DB,  and  we  have 

CB<  CD  +  DA. 
That  is,  CB<CA.  O.B.O. 


PERPENDICULAR  AND   OBLIQUE   LINES.  39 

123,  Since  two  points  determine  the  position  of  a  straight 
line,  two  points  equidistant  from  the  extremities  of  a  line  deter- 
mine  the  perpendicular  at  the  middle  of  that  line. 

THE  Locus  OF  A  POINT. 

124,  If  it  is  required  to  find  a  point  which  shall  fulfil  a 
single  geometric  condition,  the  point  will  have  an  unlimited 
number  of  positions,  but  will  be  confined  to  a  particular  liney 
or  group  of  lines. 

Thus,  if  it  is  required  to  find  a  point  equidistant  from  the 
extremities  of  a  given  straight  line,  it  is  obvious  from  the  last 
proposition  that  any  point  in  the  perpendicular  to  the  given 
line  at  its  middle  point  does  fulfil  the  condition,  and  that  no 
other  point  does ;  that  is,  the  required  point  is  confined  to  this 
perpendicular.  Again,  if  it  is  required  to  find  a  point  at  a 
given  distance  from  a  fixed  straight  line  of  indefinite  length,  it 
is  evident  that  the  point  must  lie  in  one  of  two  straight  lines, 
so  drawn  as  to  be  everywhere  at  the  given  distance  from  the 
fixed  line,  one  on  one  side  of  the  fixed  line,  and  the  other  on 
the  other  side. 

The  locus  of  a  point  under  a  given  condition  is  the  line, 
or  group  of  lines,  which  contains  all  the  points  that  fulfil  the 
given  condition,  and  no  other  points. 

125,  SCHOLIUM.   In  order  to  prove  completely  that  a  certain 
line  is  the  locus  of  a  point  under  a  given  condition,  it  is  neces- 
sary to  prove  that  every  point  in  the  line  satisfies  the  given 
condition;  and  secondly,  that  every  point  which  satisfies  the 
given  condition  lies  in  the  line  (the  converse  proposition),  or 
that  every  point  not  in  the  line  does  not  satisfy  the  given  condi- 
tion (the  opposite  proposition). 

126,  COR.   The  locus  of  a  point  equidistant  from  the  extrem- 
ities of  a  straight  line  is  the  perpendicular  bisector  of  that  line. 

§§  122,  123 


40 


PLANE   GEOMETEY.  —  BOOK  I. 


J> 


TRIANGLES. 

127.  A  triangle  is  a  portion  of  a  plane  bounded  by  three 
straight  lines;  as,  ABC. 

The  bounding  lines  are  called  the 
sides  of  the  triangle,  and  their  sum  is 
called  its  perimeter ;  the  angles  formed 
by  the  sides  are  called  the  angles  of  the 
triangle,  and  the  vertices  of  these  an- 
gles, the  vertices  of  the  triangle. 

128,  An  exterior  angle  of  a  triangle 
is  an  angle  formed  between  a  side  and 
the  prolongation  of  another  side ;  as, 

ACD.     The   interior   angle   AGE  is    

adjacent   to   the    exterior   angle  ;  the  FIG.  2. 

other  two  interior   angles,   A   and   B,  are   called   opposite- 
interior  angles. 


FIGL  1. 


Scalene. 


Isosceles. 


Equilateral. 


129,  A  triangle  is  called,  with  reference  to  its  sides,  a 
scalene  triangle  when  no  two  of  its  sides  are  equal ;  an  isos- 
celes triangle,  when  two  of  its  sides  are  equal ;  an  equilateral 
triangle,  when  its  three  sides  are  equal. 


Right. 


Obtuse. 


Acute. 


Equiangular. 


130,   A  triangle  is  called,  with  reference  to  its  angles,  a  right 
triangle,  when  one  of  its  angles  is  a  right  angle ;  an  obtuse 


TEIANGLES.  41 

triangle,  when  one  of  its  angles  is  an  obtuse  angle ;  an  acute 
triangle,  when  all  three  of  its  angles  are  acute  angles ;  an 
equiangular  triangle,  when  its  three  angles  are  equal. 

131,  In  a  right  triangle,  the  side  opposite  the  right  angle  is 
called  the  hypotenuse,  and  the  other  two  sides  the  legs,  of  the 
triangle. 

132,  The  side  on  which  a  triangle  is  supposed  to  stand  is 
called  the  base  of  the  triangle.      Any  one  of  the  sides  may  be 
taken  as  the  base.      In  the  isosceles  triangle,  the  equal  sides 
are  generally  called  the  legs,  and  the  other  side,  the  base. 

133,  The  angle  opposite  the  base  of  a  triangle  is  called  the 
vertical  angle,  and  its  vertex  the  vertex  of  the  triangle. 

134,  The  altitude  of  a  triangle  is  the  perpendicular  distance 
from  the  vertex  to  the  base,  or  to  the  base  produced ;  as,  AD. 

135,  The  three  perpendiculars  from  the  vertices  of  a  tri- 
angle to  the  opposite  sides  (produced  if  necessary)  are  called 
the  altitudes;  the  three  bisectors  of  the  angles  are  called  tha 
bisectors;  and  the  three  lines  from  the  vertices  to  the  middle 
points   of  the   opposite  sides  are  called   the  medians  of  the 
triangle. 

136,  If  two  triangles  have  the  angles  of  the  one  equal  respec- 
tively to  the  angles  of  the  other,  the  equal  angles  are  called 
homologous  angles,  and  the  sides  opposite  the  equal  angles  are 
called  homologous  sides. 

In  general,  points,  lines,  and  angles,  similarly  situated  in 
equal  or  similar  figures,  are  called  homologous. 

137,  THEOREM.    The  sum  of  two  sides  of  a  triangle  is  greater 
than  the  third  side,  and  their  difference  is  less  than  the  third 
side. 

In  the  A  ABC  (Fig.  1),  AB  +  BOAC,  for  a  straight  line 
is  the  shortest  distance  between  two  points ;  and  by  taking 
away  BO  from  both  sides,  AB>,AC-BC,  or  AO-BC<AB. 


42  PLANE   GEOMETRY.  —  BOOK   I. 


PROPOSITION  XXIII.    THEOREM. 

138,  The  sum  of  the  three  angles  of  a  triangle  is 
equal  to  two  right  angles. 


A  C  F 

Let  ABC  be  a  triangle. 

To  prove       Z  B  +  Z  BOA  +  ZA  =  2it.A. 

Proof,    Suppose  CE  drawn  II  to  AB,  and  prolong  A  O  to  F. 

Then       Z  ECF+  Z  ECB  +  Z  BOA  =  2  rt.  A,  §  92 

(the  sum  of  all  the  A  about  a  point  on  the  same  side  of  a  straight  line 

=  2  rt.  A). 

But  Z  A  =  Z  EOF,  §  106 

(being  ext.-int.  A  of  II  lines). 

and  Z£  =  Z  _£<?.#,  §104 

(being  alt.-int.  Aof\\  lines). 

Substitute  for  Z  ^OFand  Z  £<?.#  the  equal  A  A  and  ^. 
Then  /.A+Z.B  +  Z.BCA  =  2rt.A. 

Q.  E.  D. 

139,  COR.  1.  If  the  sum  of  two  angles  of  a  triangle  is  sub- 
tracted from  two  right  angles,  the  remainder  is  equal  to  the 
third  angle. 

140,  COR.  2.  If  two  triangles  have  two   angles  of  the  one 
equal  to  two  angles  of  the  other,  the  third  angles  are  equal. 

141,  COR.  3.  If  two  right  triangles  have  an  acute  angle  of 
the  one  equal  to  an  acute  angle  of  the  other ,  the  other  acute 
angles  are  equal. 


TRIANGLES.  43 

142,  COR.  4.  In  a  triangle  there  can  be  but  one  right  angle, 
or  one  obtuse  angle. 

143,  COR.  5.   In  a  right  triangle  the  two  acute  angles  are 
complements  of  each  other. 

144,  COR.  6.  In  an  equiangular  triangle,  each  angle  is  one- 
third  of  two  right  angles,  or  two-thirds  of  one  right  angle. 


PROPOSITION  XXIV.    THEOREM. 

145,  The  exterior  angle  of  a  triangle  is  equal  to  the 
sum  of  the  two  opposite  interior  angles. 


-**•  O 

Let  BCH  be  an  exterior  angle  of  the  triangle  ABC. 
To  prove  Z.BOH=< 

Proof,  Z  ECH+  Z . 

(being  sup.-adj.  A), 
(the  sum  of  the  three  A  of  a  A  =  2  rt.  A}. 

,       /     T)  /"Y  TT    I      /      A  /">  ~D /      A      \        /     ~D     I       /     A  /~V  T)  A  ,,     1 

•  •  *  _  jD  \j  J^L  ~T~  '     ^L  \JJLJ  —  x    _^j_  — j—  x      /)  — t—  x    _^JL  ( y  /j .        xi.X.  JL 

Take  away  from  each  of  these  equals  the  common  Z  .4(7.5 ; 
then  Z  ^(7^=  Z  ^4  +  Z  .5.  Ax.  3 

Q.  E.  D. 

146.   COR.    The  exterior  angle  of  a  triangle  is  greater  than 
either  of  the  opposite  interior  angles. 


44  PLANE    GEOMETRY.  —  BOOK   I. 


7      PROPOSITION  XXV.    THEOREM. 

147.  Two  triangles  are  equal  if  a  side  and  two  ad- 
jacent angles  of  the  one  are  equal  respectively  to  a 
side  and  two  adjacent  angles  of  the  other. 


A  CD 

In  the  triangles  ABC  and  DEF,  let  AB  =  DE,  £A  = 


To  prove  A  ABC=  A  DEF. 

Proof,   Apply  the  A  ABC  to  the  A  DEF  so  that  AB  shall 
coincide  with  DE. 

A  C  will  take  the  direction  of  DF, 
(for  ZA  =  ^.D,by  hyp)  ; 

the  extremity  C  of  AC  will  fall  upon  DF  or  DF  produced. 

JStfwill  take  the  direction  of  EF, 
(for  /.B  =  /LE,by  hyp.)  • 

the  extremity  C  of  BC  will  fall  upon  EF  or  EF  produced. 

/.the  point  (7,  falling  upon  both  the  lines  DF  and  EF, 
must  fall  upon  the  point  common  to  the  two  lines,  namely,  F. 

/.the  two  A  coincide,  and  are  equal.  Q.E.  D. 

148,  Con.  1.   Two  right  triangles  are  equal  if  the  hypotenuse 
and  an  acute  angle  of  the  one  are  equal  respectively  to  the  hypote- 
nuse and  an  acute  angle  of  the  other. 

149,  COR.  2.  Two  right  triangles  are  equal  if  a  side  and  an 
acute  angle  of  the  one  are  equal  respectively  to  a  side  and 
homologous  acute  angle  of  the  other, 


TRIANGLES.  45 


PROPOSITION  XXVI.     THEOREM. 

150,  Two  triangles  are  equal  if  two  sides  and  the 
included  angle  of  the  one  are  equal  respectively  to 
two  sides  and  the  included  angle  of  the  other. 


A  B  D  E 

In  the  triangles  ABC  and  DEF,  let  AB  =  DE,  AC  =  DF, 

To  prove  A  ABC = A  DEF. 

Proof,   Apply  the  A  ABC  to  the  A  DEF  so  that  AB  shall 
coincide  with  DE. 

Then  AC  will  take  the  direction  of  DF, 

(for  /.A  =  ^D,by  hyp.) ; 

the  point  C  will  fall  upon  the  point  F, 
(forAC=DFt  by  hyp.). 


(their  extremities  being  the  same  points). 
/.the  two  A  coincide,  and  are  equal. 


Q.E.D. 


151,   COR.    Two  right  triangles  are  equal  if  their  legs  are 
equal,  each  to  each. 


46 


PLANE    GEOMETRY.  —  BOOK   I. 


PROPOSITION  XXVII.     THEOREM. 

152,  If  two  triangles  have  two  sides  of  the  one  equal 
respectively  to  two  sides  of  the  other,  but  the  included 
angle  of  the  first  greater  than  the  included  angle  of 
the  second,  then  the  third  side  of  the  first  will  be 
greater  than  the  third  side  of  the  second. 


E 

In  the  triangles  ABC  and  ABE,  let  AB  =  AB,  BC=BE; 
but  /.ABO  greater  than  /.ABE. 

To  prove  AC>AE. 

Proof,    Place  the  A  so  that  AB  of  the  one  shall  coincide  with 
AB  of  the  other. 

Suppose  BF  drawn  so  as  to  bisect  Z  EBQ. 

Draw  EF. 
In  the  A  EEF  *xA  CBF 

EB  =  EC,  Hyp. 

BF=BF,  Iden. 

Z  EBF=  Z.  CBF.  Cons. 

/.  the  A  EBF  and  CEFms  equal,  §  150 

(having  two  sides  and  the  included  Z.  of  one  equal  respectively  to  two  sides 
and  the  included  /.  of  the  other). 

.-.EF=FC,  • 

(being  homologous  sides  of  equal  A). 

Now  AF+  FE  >  AE,  §  137 

(the  sum  of  two  sides  of  a  A  is  greater  than  the  third  side). 

.-.AF+FOAE-, 

or,  AC>AE.  Q.E.D. 


TRIANGLES.  47 


PROPOSITION  XXVIII.    THEOREM. 

153,  CONVERSELY.  If  two  sides  of  a  triangle  are  equal 
respectively  to  two  sides  of  another,  but  the  third  side 
of  the  first  triangle  is  greater  than  the  third  side  of 
the  second,  then  the  angle  opposite  the  third  side  of 
the  first  triangle  is  greater  than  the  angle  opposite 
the  third  side  of  the  second. 

D 

A 


B  C  E  F 

In  the  triangles  ABC  and  DEFt  let  AB  =  DE,  AC  =  DF; 
but  let  BG  be  greater  than  EF. 
To  prove  Z  A  greater  than  Z  D. 

Proof.    Now  Z  A   is  equal  to  Z  D,  or  less  than  Z  D,  or 
greater  than  Z  D. 

But  Z  A  is  not  equal  to  Z  D,  for  then  A  ABC  would  be 
equal  to  A  DEF,  §  150 

(having  two  sides  and  the  included  Z.  of  the  one  respectively  equal  to  two 
sides  and  the  included  /.  of  the  other), 

and  BC  would  be  equal  to  EF. 

And  Z  A  is  not  less  than  Z  D,  for  then  BC  would  be  less 
than  EF.  §  152 

.'.  Z  A  is  greater  than  Z  D. 

Q.E.D. 


PLANE   GEOMETRY. — BOOK   L 


PROPOSITION  XXIX.    THEOREM. 

154,  In  an  isosceles  triangle  the  angles  opposite  the 
equal  sides  are  equal. 


B  D  C 

Let  ABC  be  an  isosceles  triangle,  having  the  sides 
AB  and  AC  equal. 

To  prove  Z  B  —  Z  C. 

Proof,    Suppose  AD  drawn  so  as  to  bisect  the  Z.BAO. 
In  the  A  ADB  and  ADO, 

--AC.  Hyp. 

>  =  AD,  Iden. 

Z  BAD  =  Z  CAD.  Cons. 

(two  A  are  et/rtaZ  if  two  sides  and  the  included  Z  of  the  one  are  equal 


respectively  to  two  sides  and  the  included  Z.  of  the  other). 


Q.E.D. 


155.  COR.  An  equilateral  triangle  is  equiangular,  and  each 
angle  contains  60°. 

Ex.  14.  The  bisector  of  the  vertical  angle  of  an  isosceles  triangle 
bisects  the  base,  and  is  perpendicular  to  the  base. 

Ex.  15.  The  perpendicular  bisector  of  the  base  of  an  isosceles  triangle 
passes  through  the  vertex  and  bisects  the  angle  at  the  vertex. 


TKIANGLES.  49 


PROPOSITION  XXX.    THEOREM. 

156,  If  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  the  equal  angles  are  equal,  and  the  triangle 
is  isosceles. 


B  D  c 

In  the  triangle  ABC,  let  the  ZJ5  =  ZC. 

To  prove  AB  =  AC. 

Proof,  Suppose  AD  drawn  JL  to  BO. 

In  the  rt.  A  ALB  and  ADC, 

AD  =  AD,  Iden. 

ZJS  =  ZG  Hyp. 

.'.  rt.  A  ADB  =  rt.  A  ADC,  §  149 

(having  a  side  and  an  acute  Z  of  the  one  equal  respectively  to  a  side  and 
an  homologous  acute  Z  of  the  other). 


(being  homologous  sides  of  equal  A). 

Q.E.  D. 

157,   COR.    An  equiangular  triangle  is  also  equilateral. 


Ex.  16.  The  perpendicular  from  the  vertex  to  the  base  of  an  isosceles 
triangle  is  an  axis  of  symmetry. 


50  PLANE    GEOMETfiY. —  BOOK   I. 


PROPOSITION  XXXI.    THEOREM. 

158,  If  two  sides  of  a  triangle  are  unequal,  the  an- 
gles opposite  are  unequal,  and  the  greater  angle  is 
opposite  the  greater  side. 


C  B 

In  the  triangle  ACB  let  AB  be  greater  than  AC. 

To  prove  Z  A  CB  greater  than  Z  B. 

Proof.  Take  AE  equal  to  AC. 

Draw  EC. 
Z.AEC=Z.ACE,  §154 

(being  A  opposite  equal  sides). 

But  Z  A  EC  is  greater  than  /.  B,  §146 

(an  exterior  Z  of  a  A  is  greater  than  either  opposite  interior  Z). 

and  Z  ACB  is  greater  than  Z  A CE.  Ax.  8 

Substitute  for  Z  ACE  its  equal  Z  AEC9 
then  Z  ACB  is  greater  than  Z  AEC. 

Much  more, then, is  the  Z  ACB  greater  than  Z  B. 

Q.  E.  D. 


Ex.  17.  If  the  angles  ABC  and  ^.OT,  at  the  base  of  an  isosceles  tri- 
angle, be  bisected  by  the  straight  lines  BD,  CD,  show  that  -ZX5(7will 
be  an  isosceles  triangle. 


TRIANGLES.  51 


PROPOSITION  XXXII.     THEOREM. 

159,  CONVERSELY  :  If  two  angles  of  a  triangle  are 
unequal,  the  sides  opposite  are  unequal,  and  the 
greater  side  is  opposite  the  greater  angle. 


In  the  triangle  ACB,  let  angle  ACB  be  greater  than 
angle  B. 

To  prove  AB  >  AC. 

Proof.  Now  AB  is  equal  to  AC,  or  less  than  AC,  or  greater 
than  AC. 

But  AB  is  not  equal  to  AC,  for  then  the  Z  C  would  be 
equal  to  the  Z  B,  §  154 

(being  A  opposite  equal  sides). 

And  AB  is  not  less  than  AC,  for  then  the  Z  C  would  be 
less  than  the  Z  B,  §  158 

(if  two  sides  of  a  A  are  unequal,  the  A  opposite  are  unequal,  and  the 
greater  Z  is  opposite  the  greater  side). 

.'.  AB  is  greater  than  AC. 

Q.  E.  D. 


Ex.  18.  .45(7  and  ABD  are  two  triangles  on  the  same  base  AB,  and 
on  the  same  side  of  it,  the  vertex  of  each  triangle  being  without  the 
other.  If  AC  equal  AD,  show  that  EG  cannot  equal 
ED. 

Ex.  19.  The  sum  of  the  lines  which  join  a  point 
within  a  triangle  to  the  three  vertices  is  less  than 
the  perimeter,  but  greater  than  half  the  perimeter. 


52 


PLANE   GEOMETRY. BOOK  I. 


PROPOSITION  XXXIII.    THEOREM. 

160,  Two  triangles  are  equal  if  the  three  sides  of 
the  one  are  equal  respectively  to  the  three  sides  of 
the  other. 
B 


In  the  triangles  ABC  and  A'B'C',  let  AB  =  A'B',  AC^A'CP, 
BC=B'C'. 

To  prove  A  AJBO=  A  A'B'Q*. 

Proof.  Place  A  A'B'C'  in  the  position  AB'C,  having  its 
greatest  side  A'C'  in  coincidence  with  its  equal  AC,  and  its 
vertex  at  £',  opposite  B  ;  and  draw  BB'. 

Hyp. 

§154 
(in  an  isosceles  A  the  A  opposite  the  equal  sides  are  equal). 

Since  CB  =  CB\  Hyp. 

Z  CBB'  =  Z  CB'B,  §  154 

Hence,  Z  ABO=  Z  AB'C,  Ax.  2 


Since  AB  =  AB\ 


=&  A£'C=  A  A'ffQ'  §  150 

(two  &  are  equal  if  two  sides  and  included  Z  of  one  are  equal  to  two 
sides  and  included  Z  of  the  other). 


Q.E.D. 


TRIANGLES. 


53 


PKOPOSITION  XXXIV.    THEOREM. 

161.  Two  right  triangles  are  equal  if  a  side  and 
the  hypotenuse  of  the  one  are  equal  respectively  to  a 
side  and  the  hypotenuse  of  the  other. 


In  the  right  triangles  ABC  and  A'B'C',  let  AB  = 
and  AC  =  A' C1. 

Proof,  Apply  the  A  ABC  to  the  A  A'ffC',  so  that  AB  shall 
coincide  with  A'B',  A  falling  upon  A',  B  upon  £',  and  C  and 
C1  upon  the  same  side  of  A'J3f. 

Then          BO  will  take  the  direction  of  B1  C9, 

(for  Z  ABC=  ^  A'B'C',  each  being  a  rt.  Z). 
Since  AC=  A'C', 

the  point  C  will  fall  upon  C",  §  121 

(two  equal  oblique  lines  from  a  point  in  a  JL  cut  off  equal  distances  from 
the  foot  of  the  JL). 


.*.  the  two  A  coincide,  and  are  equal. 


Q.E.  D. 


54  PLANE   GEOMETRY. BOOK   I. 


PROPOSITION  XXXV.     THEOREM. 

162,  Every  point  in  the  bisector  of  an  angle  is  equi- 
distant from  the  sides  of  the  angle. 


Let  AD  be  the  bisector  of  the  angle  BAG,  and  let  O 
be  any  point  in  AD. 

To  prove  that  0  is  equidistant  from  AB  and  AC. 

Proof.   Draw  O^and  OG  J-  to  AB  and  A  O  respectively. 

In  the  rt.  A  AOF&nd  AOG 

AO  =  AO,  Hen. 

£JBAO  =  Z.CAO.  Hyp. 

.:AAOF=AAOQ,  §  148 

(two  rt.  &  are  equal  if  the  hypotenuse  and  an  acute  Z.  of  the  one  are  equal 
respectively  to  the  hypotenuse  and  an  acute  Z  of  the  other). 


Q.E.D. 


=  OG, 

(homologous  sides  of  equal  &). 
/.  0  is  equidistant  from  AB  and  .401 


What  is  the  locus  of  a  point : 

Ex.  20.  At  a  given  distance  from  a  fixed  point  ?     \  57. 

Ex.21.  Equidistant  from  two  fixed  points?     §119. 

Ex.  22.  At  a  given  distance  from  a  fixed  straight  line  of  indefinite 
length  ? 

Ex.  23.  Equidistant  from  two  given  parallel  lines  ? 

EX.  24.  Equidistant  from  the  extremities  of  a  given  line  ? 


TRIANGLES.  55 


PROPOSITION  XXXVI.    THEOREM. 

163,  Every  point  within  an  angle,  and  equidistant 
from  its  sides,  is  in  the  bisector  of  the  angle. 


Let  0  be  equidistant  from  the  sides  of  the  angle 
BAG,  and  let  AO  join  the  vertex  A  and  the  point  0. 

To  prove  that  AO  is  the  bisector  of  Z  BAC. 

Proof,   Suppose  OF  and  OG  drawn  J.  to  AB  and  AC, 

respectively. 

In  the  rt.  A  ^40.Fand  AOff 

OF=  Off,  Hyp. 

AO=AO.  Iden. 

.-.  A  AOF=A  AOG,  §161 

(two  rt.  &  are  equalif  the  hypotenuse  and  a  side  of  the  one  are  equal  to  the 
hypotenuse  and  a  side  of  the  other). 


=  Z.  GAO, 

(homologous  A  of  equal  &). 
.'.  AO  is  the  bisector  of  Z  BAC. 


a  E.  D. 


164.   COR.   The  locus  of  a  point  within  an  angle,  and  equi- 
distant from  its  sides,  is  the  bisector  of  the  angle. 


56  PLANE   GEOMETRY.  —  BOOK   I. 


QUADRILATERALS. 

165,  A  quadrilateral  is  a  portion  of  a  plane  bounded  by 
four  straight  lines. 

The  bounding  lines  are  the  sides,  the  angles  formed  by  these 
sides  are  the  angles,  and  the  vertices  of  these  angles  are  the 
vertices, of  the  quadrilateral. 

166,  A  trapezium  is  a  quadrilateral  which  has  no  two  sides 
parallel. 

167,  A  trapezoid  is  a  quadrilateral  which  has  two  sides,  and 
only  two  sides,  parallel. 

168,  A  parallelogram  is  a  quadrilateral  which  has  its  oppo- 
site sides  parallel. 


Trapezium.  Trapezoid.  Parallelogram. 

169,  A  rectangle  is  a  parallelogram  which  has  its  angles 
right  angles. 

170,  A  rhomboid  is  a  parallelogram  which  has  its  angles 
oblique  angles. 

171,  A  square  is  a  rectangle  which  has  its  sides  equal. 

172,  A  rhombus  is  a  rhomboid  which  has  its  sides  equal. 


Square.  Rectangle.  Rhombus.  Rhomboid. 

173,    The  side  upon  which  a  parallelogram  stands,  and  the 
opposite  side,  are  called  its  lower  and  upper  bases. 


QUADRILATERALS.  57 

174,  The  parallel  sides  of  a  trapezoid  are  called  its  bases, 
the  other  two  sides  its  legs,  and  the  line  joining  the  middle 
points  of  the  legs  is  called  the  median. 

175,  A  trapezoid  is  called  an  isosceles  trapezoid  when  its 
legs  are  equal. 

176,  The  altitude  of  a  parallelogram  or  trapezoid  is  the 
perpendicular  distance  between  its  bases. 

177,  The  diagonal  of  a  quadrilateral  is  a 
straight  line  joining  two  opposite  vertices. 


PROPOSITION  XXXVII.     THEOREM. 

178,  The  diagonal  of  a  parallelogram  divides  the 
figure  into  two  equal  triangles. 


Let  ABCE  be  a  parallelogram  and  AC  its  diagonal. 

To  prove  A  ABC = A  AEC. 

In  the  A  ABO  and  AEC, 

AC=AC,  Iden. 

Z.  ACB  =  Z.  CAE,  §104 

and  Z.CAB  =  /.ACE, 

(being  alt. -int.  A  of  II  lines). 

.-.AA£C=AAE&,  §147 

(having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and 
two  adj.  A  of  the  other). 

Q.  E.  D. 


58  PLANE    GEOMETRY. BOOK   I. 


PROPOSITION  XXXVIII.     THEOREM. 

179,  In  a  parallelogram  the  opposite  sides  are  equal, 
and  the  opposite  angles  are  equal. 


0 


A  E 

Let  the  figure  ABCE  be  a  parallelogram. 

To  prove  BO=  AE,  and  AB  =  EC, 

also,  Z  B  =  Z  E,  and  Z  BAE=  Z  .SOS 

Proof.  Draw  AC. 

AAJBC—AAEC,  §178 

(£fo  diagonal  of  a  EU  divides  the  figure  into  two  equal  &). 

/.  B0=  AE,  and  AB  =  CE, 
(being  homologous  sides  of  equal  &). 

A  i  /    7? /    777  j      y    "DA  77* /    7?  /~Y  77*  R.    1  1 0 

xxlSO,  Z-  .13  =z  Z-  -tL/,  and  Z.  JD^LJI/  =  Z~  _r>L>JtLi,  §  11Z 


(having  their  sides  II  and  extending  in  opposite  directions  from 
their  vertices). 

Q.E.  D. 

180,  COR.    1.   Parallel  lines  comprehended  between  parallel 
lines  are  equal.  ^ B 

181.  COR.  2.  Two  parallel  lines 

are    everywhere    equally    distant.  

For  if  AB  and  DC  are  parallel,  "  D 
Js  dropped  from  any  points  in  AB  to  DC,  measure  the  distances 
of  these  points  from  DC.  But  these  J§  are  equal,  by  §  180 ; 
hence,  all  points  in  AB  are  equidistant  from  DC. 


QUADBILATEKALS.  59 


PROPOSITION  XXXIX.     THEOREM. 

182,  If  two  sides  of  a  quadrilateral  are  equal  and 
parallel,  then  the  other  two  sides  are  equal  and  par- 
allel, and  the  figure  is  a  parallelogram. 

B  C 


Let  the  figure  ABCE  "be  a  quadrilateral,  having  the 
side  AE  equal  and  parallel  to  BG. 

To  prove  AB  equal  and  II  to  EO. 

Proof,  Draw  AC. 

In  the  A  ABC  and  AEO 

BC=  AE,  Hyp. 

AC=AC,     .  Iden. 

ZBCA  =  ZCAE,  §104 

(being  alt.-int.  A  of  II  lines). 

.-.AABC=AACE,  §150 

(having  two  sides  and  the  included  Z.  of  the  one  equal  respectively  to  two 
sides  and  the  included  /.  of  the  other).        \ 

...  AB  =  EO, 

(being  homologous  sides  of  equal  &). 

Also,  ZBAO=ZACE, 

(being  homologous  A  of  equal  A). 

/.  AB  is  II  to  EC,  §  105 

(when  two  straight  lines  are  cut  by  a  third  straight  line,  if  the  alt.-int.  A 
are  equal,  the  lines  are  parallel). 

/.  the  figure  ABCE  is  a  O,  §  168 

(the  opposite  sides  being  parallel).  a  E.  a 


60  PLANE   GEOMETRY. — BOOK   I. 


PROPOSITION  XL.    THEOREM. 

183,  If  the  opposite  sides  of  a  quadrilateral  are 
equal,  the  figure  is  a  parallelogram. 

a 


Let  the  figure  ABCE  be  a  quadrilateral  having  BC= 
AE  and  AB  =  EC. 

To  prove  figure  ABCE  a  O. 
Proof.  Draw  AO. 

In  the  A  ABCa,n&  AEO 

BC=  AE,  Hyp. 

AB  =  CE,  Hyp. 

AO=  AO.  Iden. 

.-.  A  ASG=  A  AEO,  §  160 

(having  three  sides  of  the  one  equal  respectively  to  three  sides  of  the  other). 


and  £BAC=Z.ACE, 

(being  homologous  A  of  equal  &). 

.•.jB<7is  II  toAE, 

and  AB  is  II  to  EC,  §  105 

(when  two  straight  lines  lying  in  the  same  plane  are  cut  by  a  third  straight 
line,  if  the  alt.-int.  A  are  equal,  the  lines  are  parallel). 

.'.  the  figure  ABCE  is  a  O,  §  168 

(having  its  opposite  sides  parallel). 

Q.E.  D. 


QUADRILATERALS.  61 


PROPOSITION  XLI.     THEOREM. 

184,   The  diagonals  of  a  parallelogram  bisect  each 
other. 

B  a 


Let  the  figure  ABCE  be  a  parallelogram,  and  let 
the  diagonals  AC  and  BE  cut  each  other  at  0. 

To  prove  AO=OCt  and  BO  =  OK 

In  the  A  AOE  and  BOC 

AE^BC,  §179 

(being  opposite  sides  of  a  CJ). 

Z.OAE='/.OCB,  §104 

and  Z  OEA  =  /.  QBC, 

(being  alt.-int.  A  of  II  lines). 

.-.AAOE  =  A£OCt  §147 

(having  a  side  and  two  adj.  A  of  the  one  equal  respectively  to  a  side  and 
two  adj.  A  of  the  other). 


.'.  AO  =  0(7,  and  BO  =  OE, 
(being  homologous  sides  of  equal  A). 


Q.  E.  D. 


Ex.  25.   If  the  diagonals  of  a  quadrilateral  bisect  each  other,  the  figure 
is  a  parallelogram. 

Ex.  26.   The  diagonals  of  a  rectangle  are  equal. 

Ex.  27.   If  the  diagonals  of  a  parallelogram  are 
equal,  the  figure  is  a  rectangle. 

Ex.  28.   The  diagonals  of  a  rhombus  are  perpendicular  to  each  other, 
and  bisect  the  angles  of  the  rhombus. 

Ex.  29.   The  diagonals  of  a  square  are  perpendicular  to  each  other, 
and  bisect  the  angles  of  the  square. 


62  PLANE   GEOMETRY.  —  BOOK   I. 


PROPOSITION  XLIL     THEOREM. 

185.  Two  parallelograms,  having  two  sides  and  the 
included  angle  of  the  one  equal  respectively  to  two 
sides  and  the  included  angle  of  the  other,  are  equal. 

B 0  B' c' 


A  D  A 

In   the  parallelograms  ABCD  and  A'B'C'D',  let  AB  = 
A'B',  AD  =  A'D't  and  £A  =  /.A>. 

To  prove  that  the  UJ  are  equal. 

Apply  a  ABCD  to  O  A'£'C'Df,  so  that  AD  will  fall  on 
and  coincide  with  A(D'. 

Then  AB  will  fall  on  A'B\ 

(for  /.A  =  £A',by  hyp-). 
and  the  point  B  will  fall  on  B', 


Now,  BO  and  B'C'  are  both   II   to  A'D'  and  are  drawn 
through  point  Br. 

/.  the  lines  SO  and  B'Cf  coincide,  §  101 

and  O  falls  on  £'C'  or  £'C'  produced. 
In  like  manner,  D(7and  D(Cf  are  II  to  AB1  and  are  drawn 
through  the  point  Dr. 

/.  £><7and  D'C1  coincide.  §  101 

/.  the  point  O  falls  on  £'C't  or  D'Cf  produced. 

/.  O  falls  on  both  B'G'  and  D'C'. 

.'.  C  must  fall  on  the  point  common  to  both,  namely,  Or. 
.'.  the  two  UJ  coincide,  and  are  equal. 

Q.  E.  D. 

186.    COR.     Two  rectangles  having  equal  bases  and  equal 
altitudes  are  equal. 


QUADRILATERALS. 


63 


PROPOSITION  XLIIL     THEOREM. 

187,  If  three  or  more  parallels  intercept  equal  parts 
on  any  transversal,  they  intercept  equal  parts  on 
every  transversal. 


D/ 


Q 


Let  the  parallels  AH,  BK,  CM,  DP  intercept  equal 
parts  HK,  KM,  MP  on  the  transversal  HP. 

To  prove  that  they  intercept  equal  parts  AB,  BC,  CD  on  the 
transversal  AD. 

Proof,    From  A,  B,  and  (7  suppose  AE,  BF,  and  CO  drawn 
II  to  HP. 

Then  AE=  HK,  BF=  KM,  CG  =  MP,        §  180 
(parallels  comprehended  between  parallels  are  equal). 

.:AE=BF=CG.  Ax.  1 

Also  Z.A=^B  =  Z.C,  §106 

(being  ext.-int.  Aof\\  lines) ; 

and  ZE=ZF=ZG,  §112 

(having  their  sides  II  and  directed  the  same  way  from  the  vertices). 

.-.  A  ABE=  A  BCF=  A  CDG,  §  147 

(each  having  a  side  and  two  adj.  A  respectively  equal  to  a  side  and  two 
adj.  A  of  the  others). 

/.  AB  =  BC=  CD, 

(homologous  sides  of  equal  &).  *  fc  Ot 


64  PLANE   GEOMETRY.  —  BOOK   I. 

188,  COR.  1.   The  line  parallel  to  the  base  of  a  triangle  and 
bisecting  one  side  bisects  the  other  side  also. 

For,  let  DE  be  II  to  EC  and  bisect  AB. 
Draw  through  A  a  line  11  to  BC.  Then 
this  line  is  II  to  DE,  by  §  111.  The  three 

parallels   by  hypothesis   intercept   equal     ^_ 

parts  on  the  transversal  AB,  and  there-   B 

fore,  by  §187,  they  intercept  equal  parts  on  the  transversal 

AC]  that  is,  the  line  DE  bisects  AC. 

189,  COR.  2.   The  line  which  joins  the  middle  points  of  two 
sides  of  a  triangle  is  parallel  to  the  third  side,  and  is  equal  to 
half  the  third  side.    For,  a  line  drawn  through  D,  the  middle 
point  of  AB,  II  to  BO,  passes  through  E,  the  middle  point  of 
AC,  by  §  188.     Therefore,  the  line  joining  D  and  ^coincides 
with  this  parallel  and  is  11  to  BC.     Also,  since  EF  drawn  II 
to  AB  bisects  AC,  it  bisects  BC,  by  §  188 ;  that  is,  BF=  FC 
—  \  BC.    But  BDEF  is  a  O  by  construction,  and  therefore 
DE=BF=\BC. 

190,  COR.  3.  The  line  which  is  parallel  to  the  bases  of  a  trap- 
ezoid  and  bisects  one  leg  of  the  trap- 

ezoid  bisects  the  other  leg  also.     For 


ezoia  oisecis  ine  oirwr  Ley  aoso.      r  or  r 

if  parallels  intercept  equal  parts  on        / 
any  -transversal,  they  intercept  equal 
parts  on  every  transversal  by  §  187. 


A  ;B 

191,    COR.   4.     The  median   of  a 

trapezoid  is  parallel  to  the  bases,  and  is  equal  to  half  the  sum 
of  the  bases.  For,  draw  the  diagonal  DB.  In  the  A  ADB 
join  E,  the  middle  point  of  AD,  to  F,  the  middle  point  of  DB. 
Then,  by  §  189,  EF  is  II  to  AB  &n&  =  %AB.  In  the  &DBC 
join  Fto  G,  the  middle  point  of  BC.  Then  FG  is  II  to  DC 
and  =  \DC.  AB  and  FG,  being  II  to  DC,  are  11  to  each  other. 
But  only  one  line  can  be  drawn  through  F II  to  AB.  There- 
fore FG  is  the  prolongation  of  EF.  Hence  EFG  is  II  to  AB 
and  DC,  and  =  }  (AB  +  DC). 


EXERCISES.  65 


EXERCISES. 

30.  The  bisectors  of  the  angles  of  a  triangle  meet  in  a  point  which  is 
equidistant  from  the  sides  of  the  triangle. 

HINT.  Let  the  bisectors  AD  and  BE  intersect  at  0. 
Then  0  being  in  AD  is  equidistant  from  AC  and  AB. 
(Why  ?)  And  0  being  in  BE  is  equidistant  from  BO 
and  AB.  Hence  0  is  equidistant  from  AC  and  BO, 
and  therefore  is  in  the  bisector  OF.  (Why  ?) 

31.  The  perpendicular  bisectors  of  the  sides  of  a  triangle  meet  in  a 
point  which  is   equidistant  from  the  vertices  of  the 

triangle. 

HINT.  Let  the  JL  bisectors  EE/  and  DD'  intersect 
at  0.     Then  0  being  in  EE'  is  equidistant  from  A     . 
and  C.    (Why  ?)     And  O  being  in  DD/  is  equidistant  Jf 

from  A  and  B.     Hence  0  is  equidistant  from  B  and  C,  and  therefore 
is  in  the  JL  bisector  FF'.   (Why  ?) 

32.  The  perpendiculars  from  the  vertices  of  a*  triangle  to  the  opposite 
sides  meet  in  a  point. 

HINT.  Let  the  _k  be  AH,  BP,  and  CK 
Through  A,  B,  0  suppose  B'C',  AfQ',  A/B/ 
drawn  II  to  BO,  AC,  AB,  respectively.  Then 
AH  is  JL  to  B'V.  (Why?)  Now  ABCB'  and 
ACB(y  are  &  (why?),  and  AB'  =  BC,  and  A& 
=  BO.  (Why  ?)  That  is,  A  is  the  middle  point  of  B'W.  In  the  same  way, 
B  and  C  are  the  middle  points  of  A/C/  and  A/B/,  respectively.  There- 
fore, AH,  BP,  and  OS' are  the  JL  bisectors  of  the  sides  of  the  A  A'B'W. 
Hence  they  meet  in  a  point.  (Why  ?) 

33.  The  medians  of  a  triangle  meet  in  a  point  which  is  two-thirds  of 
the  distance  from  each  vertex  to  the  middle  of  the  opposite  side. 

HINT.  Let  the  two  medians  AD  and  CE  meet  in  0. 
Take  .Fthe  middle  point  of  OA,  and  G  of  00.  Join 
GF,  FE,  ED,  and  DG.  In  A  AOC,  GF  is  II  to  AC 
and  equal  to  J  AC.  (Why  ?)  DE  is  II  to  AC  and  equal 
to  $  AC.  (Why?)  Hence  DGFE  is  a  CJ.  (Why?) 
Hence  AF=  FO  ~  OD,  and  CG  =  GO  =  OE.  (Why  ?)  A  E  B 
Hence,  any  median  cuts  off  on  any  other  median  two-thirds  of  the  dis- 
tance from  the  vertex  to  the  middle  of  the  opposite  side.  Therefore  the 
median  from  B  will  cut  off  AO,  two-thirds  of  AD\  that  is,  will  pass 
through  0. 


66 


PLANE  GEOMETRY.  —  BOOK  I. 


POLYGONS  IN  GENERAL. 

192,  A  polygon  is  a  plane  figure  bounded  by  straight  lines. 
The  bounding  lines  are  the  sides  of  the  polygon,  and  their 

sum  is  the  perimeter  of  the  polygon. 

The  angles  which  the  adjacent  sides  make  with  each  other 
are  the  angles  of  the  polygon,  and  their  vertices  are  the  ver- 
tices of  the  polygon. 

The  number  of  sides  of  a  polygon  is  evidently  equal  to  the 
number  of  its  angles. 

193,  A  diagonal  of  a  polygon  is  a  line  joining  the  vertices 
of  two  angles  not  adjacent ;  as  AC,  Fig.  1. 

B 


FIG.  3. 


194,  An  equilateral  polygon  is  a  polygon  which  has  all  its 
sides  equal. 

195,  An  equiangular  polygon  is  a  polygon  which  has  all  its 
angles  equal. 

196,  A  convex  polygon  is  a  polygon  of  which  no  side,  when 
produced,  will  enter  the  surface  bounded  by  the  perimeter. 

197,  Each  angle  of  such  a  polygon  is  called  a  salient  angle, 
and  is  less  than  a  straight  angle. 

198,  A  concave  polygon  is  a  polygon  of  which  two  or  more 
sides,  when  produced,  will  enter  the  surface  bounded  by  the 
perimeter.     Fig.  3. 

199,  The  angle  FDE  is  called  a  re-entrant  angle,  and  is 
greater  than  a  straight  angle. 

If  the  term  polygon  is  used,  a  convex  polygon  is  meant. 


POLYGONS.  67 

200.  Two  polygons  are  equal  when  they  can  be  divided  by 
diagonals  into  the  same  number  of  triangles,  equal  each  to 
each,  and  similarly  placed  ;  for  the  polygons  can  be  applied 
to  each  other,  and  the  corresponding  triangles  will  evidently 
coincide. 

201.  Two  polygons  are  mutually  equiangular,  if  the  angles 
of  the  one  are  equal  to  the  angles  of  the  other,  each  to  each, 
when  taken  in  the  same  order.      Figs.  1  and  2. 

202,  The  equal  angles  in   mutually  equiangular  polygons 
are  called  homologous  angles  ;  and  the  sides  which  lie  between 
equal  angles  are  called  homologous  sides. 

203,  Two  polygons  are  mutually  equilateral,  if  the  sides  of 
the  one  are  equal  to  the  sides  of  the  other,  each  to  each,  when 
taken  in  the  same  order.     Figs.  1  and  2. 


FIG.  4.  FIG.  5.  FIG.  6.  FIG.  7. 

Two  polygons  may  be  mutually  equiangular  without  being 
mutually  equilateral ;  as,  Figs.  4  and  5. 

And,  except  in  the  case  of  triangles,  two  polygons  may  be 
mutually  equilateral  without  being  mutually  equiangular  ;  as, 
Figs.  6  and  7. 

If  two  polygons  are  mutually  equilateral  and  equiangular, 
they  are  equal,  for  they  may  be  applied  the  one  to  the  other 
so  as  to  coincide. 

204,  A  polygon  of  three  sides  is  called  a  trig  on  or  triangle; 
one  of  four  sides,  a  tetragon  or  quadrilateral ;  one  of  five  sides, 
&  pentagon;  one  of  six  sides,  a  hexagon;  one  of  seven  sides,  a 
heptagon;  one  of  eight  sides,  an  octagon;  one  of  ten  sides, 
decagon ;  one  of  twelve  sides,  a  dodecagon. 


a 


68 


PLANE    GEOMETRY.  —  BOOK    L 


PROPOSITION  XLIV.     THEOREM. 

205,  The  sum  of  the  interior  angles  of  a  polygon  is 
equal  to  two  right  angles,  taken  as  many  times  less 
two  as  the  figure  has  sides. 


Let  the  figure  ABCDEF  be  a  polygon  havingn  sides. 

To  prove  ZA  +  ZJ3  +  ZC,  etc.  =  (n— 2)  2  rt. A. 

Proof,  From  the  vertex  A  draw  the  diagonals  AC,  AD, 
and  AE. 

The  sum  of  the  A  of  the  A  =  the  sum  of  the  A  of  the 
polygon. 

Now  there  are  (n  —  2)  A, 

and  the  sum  of  the  A  of  each  A  =  2  rt.  A.         §  138 

.*.  the  sum  of  the  A  of  the  A,  that  is,  the  sum  of  the  A  of 
the  polygon  =  (n  —  2)  2  rt.  A.  o.  E.  a 

206.  COR.  The  sum  of  the  angles  of  a  quadrilateral  equals 
two  right  angles  taken  (4  —  2)  times,  i.e.,  equals  4  right  angles ; 
and  if  the  angles  are  all  equal,  each  angle  is  a  right  angle.  In 
general,  each  angle  of  an  equiangular  polygon  of  n  sides  is 

equal  to  — ^ <•  right  angles. 

n 


POLYGONS.  69 


PEOPOSITION  XLV.     THEOREM. 

207,  The  exterior  angles  of  a  polygon,  made  by  pro- 
ducing each  of  its  sides  in  succession,  are  together 
equal  to  four  right  angles. 


T 

Let  the  figure  ABODE  be  a  polygon,  having  its  sides 
produced  in  succession. 

To  prove  the  sum  of  the  ext.  A  =  4  rt.  A. 
Proof,    Denote  the  int.  A  of  the  polygon  by  A,  B,  C,  D,  E, 
and  the  ext.  A  by  a,  b,  c,  d,  e. 

ZA  +  Za  =  2Yt.A,  §90 

and  Z  B  +  Z  b  =  2  rt.  A, 

(being  sup.-adj.  A). 

I 

In  like  manner  each  pair  of  adj .  A  —  2  rt.  A. 

:.  the  sum  of  the  interior  and  exterior  A=2  rt.  A  taken 
as  many  times  as  the  figure  has  sides, 

or,  2  n  rt.  A. 

But  the  interior  A  —  2  rt.  A  taken  as  many  times  as  the 
figure  has  sides  less  two,  —  (n—  2)  2  rt.  A, 

or,  2  n  rt.  A  —  4  rt.  A 

.*.  the  exterior  /4  =  4  rt.  2! 

Q.E.D. 


70 


PLANE   GEOMETRY.  —  BOOK   I. 


PROPOSITION  XL VI.     THEOREM. 

208,  A  quadrilateral  which  has  two  adjacent  sides 
equal,  and  the  other  two  sides  equal,  is  symmetrical 
with  respect  to  the  diagonal  joining  the  vertices  of 
the  angles  formed  by  the  equal  sides,  and  the  diago- 
nals intersect  at  right  angles. 


Let  ABCD  be  a  quadrilateral,  having  AB  =  AD,  and 
CE-GDy  and  having  the  diagonals  AC  and  BD. 

To  prove  that  the  diagonal  AC  is  an  axis  of  symmetry,  and 
is  JL.  to  the  diagonal  BD. 

Proof,   In  the  A  ABC  &K&  ADC 

AE  =  AD,  and  BC=  DC,  Hyp. 

and  AC=AO.  Iden. 

.\A  ABC  =  A  ADC,  §160 

(having  three  sides  of  the  one  equal  to  three  sides  of  the  other). 

.-.  Z  BAC=  Z  DAG,  and  Z  BGA=/-DCA, 

(homologous  A  of  equal  A). 

Hence,  if  ABC  is  turned  on  AC  as  an  axis,  AB  will  fall 
upon  AD,  CB  on  CD,  and  OB  on  OD. 

Hence  A  G  is  an  axis  of  symmetry,  §  65,  and  is  _L  to  BD. 

Q.E.O. 


POLYGONS. 


71 


PROPOSITION  XLVII.     THEOREM. 

209,  //  a  figure  is  symmetrical  with  respect  to  two 
axes  perpendicular  to  each  other,  it  is  symmetrical 
with  respect  to  their  intersection  as  a  centre. 


Let  the  figure  ABCDEFGH  be  symmetrical  with 
respect  to  the  two  axes  XX' t  YY',  which  intersect  at 
right  angles  at  0. 

To  prove  0  the  centre  of  symmetry  of  the  figure. 
Proof,    Let  N  be  any  point  in  the  perimeter  of  the  figure. 
Draw  NMIA,  to  YY'  and  IKL  J_  to  XX*. 
Join  LO,  ON,  and  KM. 

Now  KI=  KL, 

(the  figure  being  symmetrical  with  respect  to  XX'). 

But  KI=  OM, 

(Us  comprehended  between  Us  are  equal). 

/.  KL=OM,w&  KLOMis  a  O, 

(having  two  sides  equal  and  parallel). 

.'.  LO  is  equal  a,nd  parallel  to  KM. 
In  like  manner  we  may  prove  ON  equal  and  parallel  to  KM. 
Hence  the  points  L,  0,  and  .TV  are  in  the  same  straight  line 
drawn  through  the  point  0  II  to  KM]  and  LO=ONy  since 
each  is  equal  to  KM. 
.'.  any  straight  line  LON,  drawn  through  0,  is  bisected  at  0. 

.*.  0  is  the  centre  of  symmetry  of  the  figure.         §  64 

Q.  E.  D. 


§61 
§180 
§182 
§179 


72  PLANE   GEOMETRY.  —  BOOK   I. 

EXERCISES. 

34.  The  median  from  the  vertex  to  the  base  of  an  isosceles  triangle  is 
perpendicular  to  the  base,  and  bisects  the  vertical  angle. 

35.  State  and  prove  the  converse. 

36.  The  bisector  of  an  exterior  angle  of  an  isosceles  triangle,  formed 
by  producing  one  of  the  legs  through  the  vertex,  is  parallel  to  the  base. 

37.  State  and  prove  the  converse. 

38.  The  altitudes  upon  the  legs  of  an  isosceles  triangle  are  equal. 

39.  State  and  prove  the  converse. 

40.  The  medians  drawn  to  the  legs  of  an  isosceles  triangle  are  equal. 

41.  State  and  prove  the  converse.     (See  Ex.  33.) 

42.  The  bisectors  of  the  base  angles  of  an  isosceles  triangle  are  equal. 

43.  State  the  converse  and  the  opposite  theorems. 

44:  The  perpendiculars  dropped  from  the  middle  point  of  the  base  of 
an  isosceles  triangle  upon  the  legs  are  equal. 

45.  State  and  prove  the  converse. 

^  46.  If  one  of  the  legs  of  an  isosceles  triangle  is  produced  through  the 
vertex  by  its  own  length,  the  line  joining  the  end  of  the  leg  produced  to 
the  nearer  end  of  the  base  is  perpendicular  to  the  base. 

-  47.   Show  that  the  sum  of  the  interior  angles  of  a  hexagon  is  equal  to 
eight  right  angles. 

48.  Show  that  each  angle  of  an  equiangular  pentagon  is  f  of  a  right 
angle. 

49.  How  many  sides  has  an  equiangular  polygon,  four  of  whose  angles 
are  together  equal  to  seven  right  angles  ? 

50.  How  many  sides  has  a  polygon,  the  sum  of  whose  interior  angles 
is  equal  to  the  sum  of  its  exterior  angles  ? 

51.  How  many  sides  has  a  polygon,  the  sum  of  whose  interior  angles 
is  double  that  of  its  exterior  angles  ? 

52.  How  many  sides  has  a  polygon,  the  sum  of  whose  exterior  angles 
is  double  that  of  its  interior  angles  ? 


EXERCISES.  73 

53.  BAG  is  a  triangle  having  the  angle  B  double  the  angle  A.    If  BD 
bisect  the  angle  B,  and  meet  AC  in  D,  show  that  BD  is  equal  to  AD. 

54.  If  from  any  point  in  the  base  of  an  isosceles  triangle  parallels  to 
the  legs  are  drawn,  show  that  a  parallelogram  is  formed  whose  perimeter 
is  constant,  and  equal  to  the  sum  of  the  legs  of  the  triangle. 

55.  The  lines  joining  the  middle  points  of  the  sides  of  a  triangle  divide 
the  triangle  into  four  equal  triangles. 

56.  The  lines  joining  the  middle  points  of  the  side  of  a  square,  taken 
in  order,  enclose  a  square. 

57.  The  lines  joining  the  middle  points  of  the  sides  of  a  rectangle, 
taken  in  order,  enclose  a  rhombus. 

58.  The  lines  joining  the  middle  points  of  the  sides  of  a  rhombus, 
taken  in  order,  enclose  a  rectangle. 

59.  The  lines  joining  the  middle  points  of  the  sides  of  an  isosceles 
trapezoid,  taken  in  order,  enclose  a  rhombus  or  a  square. 

60.  The  lines  joining  the  middle  points  of  the  sides  of  any  quadri- 
lateral, taken  in  order,  enclose  a  parallelogram. 

61.  The  median  of  a  trapezoid  passes  through  the  middle  points  of 
the  two  diagonals. 

62.  The  line  joining  the  middle  points  of  the  diagonals  of  a  trapezoid 
is  equal  to  half  the  difference  of  the  bases. 

63.  In  an  isosceles  trapezoid  each  base  makes         Q J2 

equal  angles  with  the  legs.  /  \  \ 

HINT.   Draw  CE  \\  DB.  I     \  \ 

64.  In  an  isosceles  trapezoid  the  opposite  angles    •*        E 
are  supplementary. 

65.  If  the  angles  at  the  base  of  a   trapezoid   are  equal,  the   other 
angles  are  equal,  and  the  trapezoid  is  isosceles. 

66.  The  diagonals  of  an  isosceles  trapezoid  are  equal. 

67.  If  the  diagonals  of  a  trapezoid  are  equal,  the 
trapezoid  is  isosceles. 

HINT.  Draw  CE  and  DF  JL  to  CD.  Show  that  A 
ADF  and  BCE  are  equal,  that  &  COD  and  AOB  are 
isosceles,  and  that  &  AOC  and  BOD  are  equal. 


74  PLANE   GEOMETRY.  —  BOOK   I. 

68.  ABCD  is  a  parallelogram,  E  and  F  the  middle  points  of  AD  and 
.Z? (7 respectively ;  show  that  .Z^and  DFvi\\\  trisect  the  diagonal  AC. 

69.  If  from  the  diagonal  BD  of  a  square  ABCD,  BE  is  cut  off  equal 
to  BC,  and  EF  is  drawn  perpendicular  to  BD  to  meet  DC  at  F,  show 
that  DE  is  equal  to  EF,  and  also  to  FC. 

70.  The  bisector  of  the  vertical  angle  A  of  a  triangle  ABC,  and  the 
bisectors  of  the  exterior  angles  at  the  base  formed  by  producing  the  sides 
AB  and  AC,  meet  in  a  point  which  is  equidistant  from  the  base  and  the 
sides  produced. 

71.  If  the  two  angles  at  the  base  of  a  triangle  are  bisected,  and 
through  the  point  of  meeting  of  the  bisectors  a  line  is  drawn  parallel  to 
the  base,  the  length  of  this  parallel  between  the  sides  is  equal  to  the  sum 
of  the  segments  of  the  sides  between  the  parallel  and  the  base. 

72.  If  one  of  the  acute  angles  of  a  right  triangle  is  double  the  other, 
the  hypotenuse  is  double  the  shortest  side. 

73.  The  sum  of  the  perpendiculars  dropped  from  any  point  in  the 
base  of  an  isosceles  triangle  to  the  legs  is  constant, 

and  equal  to  the  altitude  upon  one  of  the  legs. 

HINT.  Let  PD  and  PE  be  the  two  Js,  BF  the  F£ 

altitude  upon  AC.    .Draw  PQ  _L  to  BF,  and  prove 

the  A  PBQ  and  PBD  equal.  ^g 

*        P 

74.  The  sum  of  the  perpendiculars  dropped  from  any  point  within  an 
equilateral  triangle  to  the  three  sides  is  constant,  and  equal   to  the 
altitude. 

HINT.  Draw  through  the  point  a  line  II  to  the  base,  and  apply  Ex.  73. 

75.  What  is  the  locus  of  all  points  equidistant  from  a  pair  of  inter- 
eecting  lines  ? 

76.  In  the  triangle  CAB  the  bisector  of  the  angle  C  makes  with  the 
perpendicular  from  C  to  AB  an  angle  equal  to  half  the  difference  of  the 
angles  A  and  B. 

77.  If  one  angle  of  an  isosceles  triangle  is  equal  to  60°,  the  triangle 
is  equilateral. 


BOOK   II. 
THE     CIRCLE. 


DEFINITIONS. 

210,  A  circle  is  a  portion  of  a  plane  bounded  by  a  curved 
line  called  a  circumference,  all  points  of  which  are  equally  dis- 
tant from  a  point  within  called  the  centre. 

211,  A  radius  is  a  straight  line  drawn  from  the  centre  to  the 
circumference ;  and  a  diameter  is  a  straight  line  drawn  through 
the  centre,  having  its  extremities  in  the  circumference. 

By  the  definition  of  a  circle,  all  its  radii  are  equal.     All  its 
diameters  are  equal,  since  the  diameter  is  equal  to  two  radii. 

212,  A  secant  is  a  straight  line  which  intersects  the  circum- 
ference in  two  points  ;  as,  AD,  Fig.  1. 

213,  A  tangent  is  a  straight  line  which  touches  the  circum- 
ference but  does  not  intersect  it;    as, 

-5(7,  Fig.  1.  The  point  in  which  the 
tangent  touches  the  circumference  is 
called  the  point  of  contact,  or  point  of 
iangency. 

214,  Two  circumferences  are  tangent 

to  each  other  when  they  are  both  tan-  Fia-  li 

gent  to  a  straight  line  at  the  same  point;  and  are  tangent 
internally  or  externally,  according  as  one  circumference  lies 
wholly  within  or  without  the  other. 


76  PLANE   GEOMETRY. BOOK    II. 

215,  An  arc  of  a  circle  is  any  portion  of  the  circumference. 
An  arc  equal  to  one-half  the  circumference  is  called  a  semi- 
circumference. 

216.  A  chord  is  a  straight  line  having  its  extremities  in  the 
circumference. 

Every  chord  subtends  two  arcs  whose  sum  is  the  circum- 
ference ;  thus,  the  chord  AB  (Fig.  3)  subtends  the  smaller  arc 
AB  and  the  larger  arc  BODE  A.  If  a  chord  and  its  arc  are 
spoken  of,  the  less  arc  is  meant  unless  it  is  otherwise  stated. 


217,  A  segment  of  a  circle  is  a  portion  of  a  circle  bounded 
by  an  arc  and  its  chord. 

A  segment  equal  to  one-half  the  circle  is  called  a  semicircle. 

218,  A  sector  of  a  circle  is  a  portion  of  the  circle  bounded 
by  two  radii  and  the  arc  which  they  intercept. 

A  sector  equal  to  one-fourth  of  the  circle  is  called  a  quadrant. 

219,  A  straight  line  is  inscribed  in  a  circle  if  it  is  a  chord. 

220,  An  angle  is  inscribed  in  a  circle  if  its  vertex  is  in  the 
circumference  and  its  sides  are  chords. 

221,  An  angle  is  inscribed  in  a  segment  if  its  vertex  is  on 
the  arc  of  the  segment  and  its  sides  pass  through  the  extrem- 
ities of  the  arc. 

222,  A  polygon  is  inscribed  in   a   circle  if  its  sides   are 
chords  of  the  circle. 

223,  A  circle  is  inscribed  in  a  polygon  if  the  circumference 
touches  the  sides  of  the  polygon  but  does  not  intersect  thera. 


ARCS   AND   CHORDS.  77 

224,  A  polygon   is   circumscribed  about  a  circle  if  all  the 
sides  of  the  polygon  are  tangents  to  the  circle. 

225,  A  circle  is  circumscribed  about  a  polygon  if  the  circum- 
ference passes  through  all  the  vertices  of  the  polygon. 

226,  Two  circles  are  equal  if  they  have  equal  radii ;   for 
they  will  coincide  if  one  is  applied  to  the  other;  conversely, 
two  equal  circles  have  equal  radii. 

Two  circles  are  concentric  if  they  have  the  same  centre. 

PROPOSITION  I.     THEOREM. 

227,  The  diameter  of  a  circle  is  greater  than  any 
other  chord;  and  bisects  the  circle  and  the  circum- 
ference. 

M 


P 

Let  AB  be  the  diameter  of  the  circle  AMBP,  and 
AE  any  other  chord. 

To  prove  AB  >  AE,  and  AB  bisects  the  circle  and  the 
circumference. 

Proof,    I.    From  C,  the  centre  of  the  O,  draw  OH. 
CE=CB, 

(being  radii  of  the  same  circle). 

But  AC+CE>AE,  §137 

(the  sum  of  two  sides  of  a  A  is  >  the  third  side). 

Then  AO+ OB  >  AE,  or  AB  >  AE.  Ax.  9 

II.  Fold  over  the  segment  A  MB  on  AB  as  an  axis  until  it 
falls  upon  APB,  §  59.  The  points  A  and  B  will  remain  fixed; 
therefore  the  arc  A  MB  will  coincide  with  the  arc  APB ; 
because  all  points  in  each  are  equally  distant  from  the 
centre  0.  §  210 

Hence  the  two  figures  coincide  throughout  and  are  equal.  §  59 

Q.E.D. 


78  PLANE  GEOMETKY.  —  BOOK   II. 


PROPOSITION  II.    THEOREM. 

228,   A  straight  line  cannot  intersect  the  circumir 
ference  of  a  circle  in  more  than  two  points. 


LetHKbe  any  line  cutting  the  circumference  AMP. 

To  prove  that  HK  can  intersect  the  circumference  in  only  two 
points. 

Proof,  If  possible,  let  UK  intersect  the  circumference  in 
three  points  If,  P,  and  K. 

From  0,  the  centre  of  the  O,  draw  OH,  OP,  and  OK 

Then  OH,  OP,  and  OJTare  equal, 

(being  radii  of  the  same  circle). 

Hence,  we  have  three  equal  straight  lines  OH,  OP,  and  OK 
drawn  from  the  same  point  to  a  given  straight  line.  But  this 
is  impossible,  §  120 

(only  two  equal  straight  lines  can  be  drawn  from  a  point  to  a  straight  line). 

Therefore,  HK  can  intersect  the  circumference  in  only  two 
points.  aE>Dt 


ARCS   AND   CHORDS.  79 


PROPOSITION  III.     THEOREM. 

•V^ 

229,  In  the  same  circle,  or  equal  circles,  equal  an- 
gles at  the  centre  intercept  equal  arcs;  CONVERSELY, 
equal  arcs  subtend  equal  angles  at  the  centre. 


p  P' 

In  the  equal  circles  ABP  and  A'B'P'  let  Z  O-=  Z  a. 
To  prove  arc  RS  =  arc  R'S1. 

Proof.  Apply  O  ABP  to  O  A'B'P, 

so  that  Z  0  shall  coincide  with  Z  O1. 

R  will  fall  upon  E\  and  8  upon  8',  §  226 

(for  OR  =  O'R',  and  08=  0'#',  being  radii  of  equal  ©). 

Then  the  arc  US  will  coincide  with  the  arc 


since  all  points  in  the  arcs  are  equidistant  from  the  centre. 

§210 
.'.arc  J2#=arc  IZ'ff. 

CONVERSELY  :     Let  arc  RS  =  arc  ReSf. 
To  prove  ZO  =  ZO'. 

Proof,  Apply  O  ABP  to  O  A'3'P1,  so  that  arc  fiS  shall  fall 
upon  arc  ^'$',  R  falling  upon  R  ',  8  upon  81,  and  0  upon  Of. 

Then  RO  will  coincide  with  Rl<J,  and  80  with  S'ff. 

,'.<£  0  and  O1  coincide  and  are  equal.  Q.  e.  a 


80  PLANE    GEOMETRY. BOOK    II. 

PROPOSITION  IV.   THEOREM. 

230,  In  the  same  circle,  or  equal  circles,  if  two 
chords  are  equal,  the  arcs  which  they  subtend  are 
equal;  CONVERSELY,  if  two  arcs  are  equal >  the  chords 
which  subtend  them,  are  equal. 


p  P' 

In  the  equal  circles  ABP  and  A'B'P',  let  chord  RS  = 
chord  R'S'. 

To  prove  arc  RS  —  arc  R'S'. 

Proof.       Draw  the  radii  OR,  08,  021',  and  08'. 

In  the  A  OR8  and  O'fi'8' 

RS=R'S',  Hyp. 

the  radii  OR  and  08=  the  radii  O'R'  and  08'.   §  226 

.'.AS08=AJR'08lt  §160 

(i^ree  sides  of  the  one  being  equal  to  three  sides  of  the  other). 

.'.zrcRS^eircR'jS',  §229 

(in  equal  ©,  equal  A  at  the  centre  intercept  equal  arcs). 

CONVEESELY  :     Let  arc  RS  =  arc  Rf8f. 

To  prove  chord  RS  =  chord  JR'8'. 

Proof.  Z  0  =  Z  0',  §  229 

(equal  arcs  in  equal  ©  subtend  equal  A  at  the  centre), 
and  OR  and  08-=  O'R1  and  0'8',  respectively.    §  226 
.'.AOJR8=A0£'8',  §150 

(having  two  sides  equal  each  to  each  and  the  included  A  equal). 

.'.  chord  £8  =  chord  R'S'.  a  E>  D. 


ARCS   AND    CHORDS.        -  81 


PROPOSITION  V.     THEOREM. 

231,  In  the  same  circle,  or  equal  circles,  if  two  arcs 
are  unequal,  and  eaeh  is  less  than  a  semi-eireumfer~ 
enee,  the  greater  are  is  subtended  by  the  greater 
chord;  CONVERSELY,  the  greater  chord  subtends  the 
greater  are. 


In  the  circle  whose  centre  is  0,  let  the  arc  AMB  be 
greater  than  the  arc  AMF. 

To  prove       chord  AB  greater  than  chord  AF. 
Proof,  Draw  the  radii  OA,  OF,  and  OB. 

Since  Fis  between  A  and  B,  OF  will  fall  between  OA  and 
OB,  and  Z  AOB  be  greater  than  Z  AOF. 
Hence,  in  the  A  AOB  and  AOF, 

the  radii  OA  and  OB  =  the  radii  OA  and  OF, 
but  Z  AOB  is  greater  than  Z  AOF. 

.'.  AB  >  AF,  §  152 

(the  &  having  two  sides  equal  each  to  each,  but  the  included  A  unequal). 

CONVERSELY  :   Let  AB  be  greater  than  AF. 

To  prove          arc  AB  greater  than  arc  AF. 
In  the  A  AOB  and  AOF, 

OA  and  OB=  OA  and  OF  respectively. 
But  AB  is  greater  than  AF.  Hyp- 

/.  Z  AOB  is  greater  than  Z  AOF,     ^    §  153 
(the  &  having  two  sides  equal  each  to  each,  but  the  third  sides  unequal). 

.'.OB  falls  without  OF. 
.'.  arc  AB  is  greater  than  arc  AF.  Q,E.  D. 


82  PLANE   GEOMETRY.  —  BOOK   II. 


PROPOSITION  VI.    THEOREM. 

232,   The  radius  perpendicular  to  a  chord  bisects 
the  chord  and  the  arc  subtended  by  it. 

E 


M 

- — 

S 

Let  AB  be  the  chord,  and  let  the  radius  OS  be  per- 
pendicular to  AB  at  M. 

To  prove    AM=  BM,  and  arc  AS = arc  BS. 

Proof,   Draw  OA  and  OB  from  0,  the  centre  of  the  circle. 

In  the  rt.  A  0AM  and  OEM 

the  radius  0 A  =  the  radius  OB, 

and  OM  =  OH.  Hen. 

.*.  A  OAM=  A  OBM,  §161 

(having  the  hypotenuse  and  a  side  of  one  equal  to  the  hypotenuse  and  a 
side  of  the  other). 

.:  AM=  BM, 


.-.  sue  AS=wcBS,  §229 

(equal  A  at  the  centre  intercept  equal  arcs  on  the  circumference). 

Q.E.  D. 

233,  COR.  1.   The  perpendicular  erected  at  the  middle  of  a 
chord  passes  through  the  centre  of  the  circle.     For  the  centre  is 
equidistant  from  the  extremities  of  a  chord,  and  is  therefore  in 
the  perpendicular  erected  at  the  middle  of  the  chord.      §  122 

234,  COR.  2.    The  perpendicular  erected  at  the  middle  of  a 
chord  bisects  the  arcs  of  the  chord. 

235,  COR.  3.    The  locus  of  the  middle  points  of  a  system  of 
parallel  chords  is  the  diameter  perpendicular  to  them. 


ARCS   AND   CHORDS.  83 


PROPOSITION  VII.     THEOREM. 

236,  In  the  same    circle,   or    equal    circles,  equal 
chords    are  equally  distant   from  the  centre;    AND 

CONVERSELY. 


Let  AB  and  OF  be  equal  chords  of  the  circle  ABFC. 

To  prove     A  13  and  CF  equidistant  from  the  centre  0. 
Proof,  Draw  OP  _L  to  AB,  Off  A.  to  CF,  and  join  OA  and  00. 

OP  and  OH  bisect  AB  and  CF,  §  232 

(a  radius  A.  to  a  chord  bisects  it). 
Hence,  in  the  rt.  A  OP  A  and  OHC 

AP=Cff,  Ax.  7 

the  radius  OA  =  the  radius  00. 

.'.  A  OP  A  -  A  OHC,  §  161 

(having  a  side  and  hypotenuse  of  the  one  equal  to  a  side  and  hypotenuse 
of  the  other). 

.-.  OP  =  Off. 

.*.  AB  and  (TFare  equidistant  from  0. 
CONVERSELY  :  Let  OP  =  OH. 
To  prove  AB  =  OF. 

Proof,   In  the  rt.  A  OP  A  and  OHC 

the  radius  OA  =  the  radius  0(7,  and  OP=  OH.  Hyp. 

/.  A  OP  A  and  Oj£T(7aie  equal.  .  §  161 


Ax  6. 

$.1.9, 


84  PLANE   GEOMETRY.  —  BOOK    II. 


PROPOSITION  VIII.    THEOREM. 

237,  In  the  same  circle,  or  equal  circles,  if  two 
chords  are  unequal,  they  are  unequally  distant  from 
the,  centre,  and  the  greater  is  at  the  less  distance. 


In  the  circle  whose  centre  is  0,  let  the  chords  AB 
and  CD  be  unequal,  and  AB  the  greater;  and  let  OE 
and  OF  be  perpendicular  to  AB  and  CD  respectively. 

To  prove  OE  <  OF. 

Proof,    Suppose  AG  drawn  equal  to  CD,  and  OH±.  to  AG. 

Then  OH=  OF,  §  236 

(in  the  same  O  two  equal  chords  are  equidistant  from  the  centre). 

Join  EH. 

OE  and  OH  bisect  AB  and  AG,  respectively,      §  232 
(a  radius  JL  to  a  chord  bisects  it). 

Since,  by  hypothesis,  AB  is  greater  than  CD  or  its  equal  AG, 
AE,  the  half  of  AB,  is  greater  than  AH,  the  half  of  AG. 

/. the  Z  AHE  is  greater  than  the  Z  A  EH,      §  158 
(the  greater  of  two  sides  of  a  A  has  the  greater  Z.  opposite  to  it). 

Therefore,  the  Z  OHE,  the  complement  of  the  Z  A  HE,  is 
less  than  the  Z  OEH,  the  complement  of  the  Z  AEH. 

.'.  OE  <  OH,  §  159 

(the  greater  of  two  AofaA  has  the  greater  side  opposite  to  it). 

S.OE<  OF,  the  equal  of  OH. 

Q.E.OL 


ARCS   AND   CHORDS.  85 


PEOPOSITION  IX.    THEOREM. 

238,  CONVERSELY  :  In  the  same  circle,  or  equal  cir- 
cles, if  two  chords  are  unequally  distant  from  the 
centre,  they  are  unequal,  and  the  chord  at  the  less 
distance  is  the  greater. 


In  the  circle  whose  centre  is  0,  let  AB  and  CD  be 
unequally  distant  from  0;  and  let  OE  perpendicular 
to  AB  be  less  than  OF  perpendicular  to  CD. 

To  prove  AB  >  CD. 

Proof,    Suppose  AG  drawn  equal  to  CD,  and  OH JL  to  AG. 

Then  OH=  OF,  §  236 

(in  the  same  O  two  equal  chords  are  equidistant  from  the  centre). 

Hence,  OE  <  OH. 
Join  EH. 

In  the  A  OEHihe  Z  OHE  is  less  than  the  Z  OEH,  §  158 

(the  greater  of  two  sides  of  a  A  has  the  greater  Z  opposite  to  it). 

Therefore,  the  Z  AHE,  the  complement  of  the  Z  OHE,  is 
greater  than  the  Z  A  EH,  the  complement  of  the  Z  OEH. 

.-.  AE  >  AH,  §  159 

(the  greater  of  two  A  of  a  A  has  the  greater  side  opposite  to  it). 

But  AE=\AB,  and  AH=\AG. 
/.  AB  >  AG\  hence  AB  >  CD,  the  equal  of  AG. 


86  PLANE   GEOMETRY.  —BOOK   II. 


PROPOSITION  X.    THEOREM. 

239,  A  straight  line  perpendicular  to  a  radius  (xb 
its  extremity  is  a  tangent  to  the  circle. 


M- 


H          A 
Let  MB  be  perpendicular  to  the  radius  OA  at  A. 

To  prove  MB  tangent  to  the  circle. 

Proof,   From  0  draw  any  other  line  to  MB,  as  OCH. 

OH>OA,  §114 

(a  _L  is  the  shortest  line  from  a  point  to  a  straight  line). 

.*.  the  point  .ZTis  without  the  circle. 

Hence,  every  point,  except  A,  of  the  line  MB  is  without  the 
circle,  and  therefore  MB  is  a  tangent  to  the  circle  at  A.  §  213 

Q.  E.  D. 

240,  COR.  1.  A  tangent  to  a  circle  is  perpendicular  to  the 
radius  drawn  to  the  point  of  contact.     For,  if  MB  is  tangent 
to  the  circle  at  A,  every  point  of  MB,  except  A,  is  without 
the  circle.    Hence,  OA  is  the  shortest  line  from  0  to  MB,  and 
is  therefore  perpendicular  to  MB  (§  114)  ;  that  is,  MB  is  per- 
pendicular to  OA. 

241,  CoK.  2.  A  perpendicular  to  a  tangent  at  the  point  of 
contact  passes  through  the  centre  of  the  circle.     For  a  radius  is 
perpendicular  to  a  tangent  at  the  point  of  contact,  and  there- 
fore, by  §  89,  a  perpendicular  erected  at  the  point  of  contact 
coincides  with  this  radius  and  passes  through  the  centre. 

242,  COR.  3.  A  perpendicular  let  fall  from  the  centre  of  a 
circle  upon  a  tangent  to  the  circle  passes  through  the  point  of 
contact. 


ARCS   AND   CHORDS. 


87 


PROPOSITION  XI.     THEOREM. 

243.  Parallels    intercept   equal   arcs   on   a  circum- 
ference. 


F' 
FIG.  1.  FIG.  2. 

Let  AB  and  CD  be  the  two  parallels. 

CASE  I.     When  AB  is  a  tangent,  and  CD  a  secant.     Fig.  1. 

Suppose  AB  touches  the  circle  at  F. 
To  prove  arc  CF=  arc  DF. 

Proof.  Suppose  FF1  drawn  J_  to  AB. 

This  J_  to  AB  at  Fis  a  diameter  of  the  circle.     §  241 
It  is  also  _L  to  CD.  §  102 

/.  arc  CF=  arc  DF,  §  232 

(a  radius  JL  to  a  chord  bisects  the  chord  and  its  subtended  arc). 

Also,  arc  FCF1  =  arc  FDF\  §  227 

.-.  arc  (FCF1  -  FC)  =±  arc  (FDF'  -  FD),       §    82 
that  is,  arc  OF1  =  arc  DF1. 

CASE  II.    When  AB  and  CD  are  secants.     Fig.  2. 
Suppose  EF  drawn  II  to  CD  and  tangent  to  the  circle  at  M. 

Then  arc  AM  —  arc  BM 

and  arc  CM    =  arc  DM-  Case  I. 

.'.by  subtraction,  arc  AC    =  arc  BD. 

CASE  III.     When  AB  and  CD  are  tangents.     Fig.  3. 
Suppose  AB  tangent  at  E,  CD  at  F,  and  GH  II  to  AB. 

Then  arc  GE    =  arc  EH  Case  I. 

and  arc  OF    —  arc  HF. 

.'.  by  addition,       arc  EGF=  arc  EHF.  Q.  E.  D. 


88  PLANE  GEOMETRY.  —  BOOK  II. 


PROPOSITION  XII.     THEOREM. 

244,    Through  three  points  not  in  a  straight  line, 
one  circumference,  and  only  one,  can  be  drawn. 


Let  A,  B,  C  be  three  points  not  in  a  straight  line. 

To  prove  that  a  circumference  can  be  drawn  through  A,  B, 
and  C,  and  only  one. 

Proof.  Join  AB  and  BO. 

At  the  middle  points  of  AB  and  B  C  suppose  Js  erected. 

Since  BC  is  not  the  prolongation  of  AB,  these  Js  will  inter- 
sect in  some  point  0. 

The  point  0,  being  in  the  J_  to  AB  at  its  middle  point,  is 

equidistant  from  A  and  B\  and  being  in  the  J.  to  BC  &i  its 

middle  point,  is  .equidistant  from  B  and  C,  §  122 

(every  point  in  the  perpendicular -bisector  of  a  straight  line  is  equidistant 

from  the  extremities  of  the  straight  line). 

Therefore  0  is  equidistant  from  A,  B,  and  C\  and  a  cir- 
cumference described  from  0  as  a  centre,  with  a  radius  OA, 
will  pass  through  the  three  given  points. 

Only  one  circumference  can  be  made  to  pass  through 
these  points.  For  the  centre  of  a  circumference  passing 
through  the  three  points  must  be  in  both  perpendiculars,  and 
hence  at  their  intersection.  As  two  straight  lines  can  inter- 
sect in  only  one  point,  0  is  the  centre  of  the  only  circumfer- 
ence that  can  pass  through  the  three  given  points.  a  E.  D< 

245,  COR.  Two  circumferences  can  intersect  in  only  two 
points.  For,  if  two  circumferences  have  three  points  common, 
they  coincide  and  form  one  circumference. 


TANGENTS,  89 

PROPOSITION  XIII.    THEOREM. 

246,  The  tangents  to  a  circle  drawn  from  an  exte- 
rior point  are  equal,  and  make  equal  angles  with 
the  line  joining  the  point  to  the  centre. 


C 

Let  AB  and  AC  be  tangents  from  A  to  the  circle 
whose  centre  is  0,  and  AO  the  line  joining  A  to  0. 

To  prove     AB  =  AC,  and  Z  BAO  =  Z  OAO. 
Proof,  Draw  OB  and  OC. 

AB  is  ±  to  OB,  and  AC  JL  to  OC,  §  240 

(a  tangent  to  a  circle  is  ±  to  the  radius  drawn  to  the  point  of  contact). 
In  the  rt.  A  OAB  and  OAO 

OB  =  OC, 

(radii  of  the  same  circle). 

OA  -  OA.  Iden. 

.'.AOAB  =  AOAC,  §161 

(having  a  side  and  hypotenuse  of  the  one  equal  to  a  side  and  hypotenuse 
of  ihz  other). 


and  Z£AO  =  Z.CAO.  Q.E.D. 

247,  DBF.  The  line  joining  the  centres  of  two  -circles  is 
called  the  line  of  centres. 

248,  DEF.  A  common  tangent  to  two  circles  is  called  a 
common  exterior  tangent  when  it  does  not  cut  the  line  of  cen- 
tres, and  a  common  interior  tangent  when  it  cuts  the  line  of 
centres. 


90  PLANE   GEOMETRY.  —  BOOK   II. 


PROPOSITION  XIV.     THEOREM. 

249.  If  two  circumferences  intersect  each  other,  the 
line  of  centres  is  perpendicular  to  their  common 
chord  at  its  middle  point. 


Let  C  and  Cf  be  the  centres  of  two  circumferences 
which  intersect  at  A  and  B.  Let  AB  be  their  common 
chord,  and  CCf  join  their  centres. 

To  prove        CC1  J_  to  AB  at  its  middle  point. 

Proof.  A  J_  drawn  through  the  middle  of  the  chord  AB 
passes  through  the  centres  C  and  C",  §  233 

(a  _L  erected  at  the  middle  of  a  chord  passes  through  the  centre  of  the  O). 

/.  the  line  CO',  having  two  points  in  common  with  this  J., 
must  coincide  with  it. 


.'.  CC'  is  J_  to  AB  at  its  middle  point. 


Q.  E.  D. 


Ex.  78.   Describe  the  relative  position  of  two  circles  if  the  line  of 
centres :  \ 

(i.)  is  greater  than  yhe  sum  of  the  radii ; 
(ii.)  is  equal  to  the  sum  of  the  radii ; 

(iii.)  is  less  than  the  sum  but  greater  than  the  difference  of  the  radii ; 
(iv.)  is  equal  to  the  difference  of  the  radii ; 
(v.)  is  less  than  the  difference  of  the  radii. 
Illustrate  each  case  by  a  figure. 


TANGENTS. 


91 


PROPOSITION  XV.     THEOREM. 

250,   If  two  circumferences  are  tangent  to  each  other, 
the  line  of  centres  passes  through  the  point  of  contact. 


Let  the  two  circumferences,  whose  centres  are  C 
and  C1,  touch  each  other  at  0,  in  the  straight  line  AB, 
and  let  CO  be  the  straight  line  joining  their  centres. 

To  prove  0  is  in  the  straight  line  OC1. 

Proof,  A  -L  to  AB,  drawn  through  the  point  0,  passes 
through  the  centres  C  and  <7f,  §  241 

(a  _L  to  a  tangent  at  the  point  of  contact  passes  through  the  centre 
of  the  circle). 

.'.  the  line  CC1,  having  two  points  in  common  with  this  _L 
must  coincide  with  it. 


/.  0  is  in  the  straight  line  CO9. 


Q.  E.  D. 


Ex.  79.   The  line  joining  the  centre  of  a  circle  to  the  middle  of  a 
chord  is  perpendicular  to  the  chord. 

Ex.  80.   The  tangents  drawn  through  the  extremities  of  a  diameter 
are  parallel. 

Ex.  81.   The  perimeter  of  an  inscribed  equilateral  triangle  is  equal 
to  half  the  perimeter  of  the  circumscribed  equilateral  triangle; 

Ex.  82.   The  sum  of  two  opposite  sides  of  a  circumscribed  quadri- 
lateral is  equal  to  the  sum  of  the  other  two  sides. 


92  PLANE   GEOMETRY.  —  BOOK   II. 


MEASUREMENT. 

251,  To  measure  a  quantity  of  any  kind  is  to  find  how  many 
times  it  contains  another  known  quantity  of  the  same  kind. 

Thus,  to  measure  a  line  is  to  find  how  many  times  it  con- 
tains another  known  line,  called  the  linear  unit. 

The  number  which  expresses  how  many  times  a  quantity 
contains  the  unit-quantity,  is  called  the  numerical  measure 
of  that  quantity  ;  as,  5  in  5  yards. 

252,  The  magnitude  of  a  quantity  is  always  relative  to  the 
magnitude  of  another  quantity  of  the  same  kind.    No  quantity 
is  great  or  small  except  by  comparison.     This  relative  magni- 
tude is  called  their  ratio,  and  is  expressed  by  the  indicated 
quotient  of  their  numerical  measures  when  the  same  unit  of 
measure  is  applied  to  both. 

The  ratio  of  a  to  b  is  written  -,  or  a  :  b. 

b 

253,  Two  quantities  that  can  be  expressed  in  integers  in 
terms  of  a  common  unit  are  said  to  be  commensurable.     The 
common  unit  is  called  a  common  measure,  and  each  quantity 
is  called  a  multiple  of  this  common  measure. 

Thus,  a  common  measure  of  2|-  feet  and  3f  feet  is  -J-  of  a 
foot,  which  is  contained  15  times  in  2^-  feet,  and  22  times  in 
3-|  feet.  Hence,  2J-  feet  and  3f  feet  are  multiples  of  -j-  of  a 
foot,  2|-  feet  being  obtained  by  taking  ^  of  a  foot  15  times,  and 
3-f  feet  by  taking  £  of  a  foot  22  times. 

254,  When   two  quantities  are  incommensurable,  that  is, 
have  no  common  unit  in  terms  of  which  both  quantities  can  be 
expressed  ,in  integers,  it  is  impossible  to  find  a  fraction  that 
will  indicate  the  exact  value  of  the  ratio  of  the  given  quanti- 
ties.    It  is  possible,  however,  by  taking  the  unit  sufficiently 
small,  to  find  a  fraction  that  shall  differ  from  the  true  value 
of  the  ratio  by  as  little  as  we  please. 


RATIO.  93 

Thus,  suppose  a  and  b  to  denote  two  lines,  such  that 


a  -  -  =  V2 

b  _  b      V2' 

Now  V2=  1.41421356  .....  ,  a  value  greater  than  1.414213, 
but  less  than  1.414214. 

If,  then,  a  millionth  part  of  b  be  taken  as  the  unit,  the  value 


of  the  ratio  y  lies  between  |HHH  and  iH^fti  and  ther-e' 
b 


fore  differs  from  either  of  these  fractions  by  less  than  - 

By  carrying  the  decimal  farther,  a  fraction  may  be  found 
that  will  differ  from  the  true  value  of  the  ratio  by  less  than  a 
billionth,  a  trillionth,  or  any  other  assigned  value  whatever. 

Expressed  generally,  when  a  and  b  are  incommensurable, 
and  b  is  divided  into  any  integral  number  (n)  of  equal  parts, 
if  one  of  these  parts  is  contained  in  a  more  than  m  times,  but 
less  than  m  +  1  times,  then 


b      n  n 

that  is,  the  value  of  2  lies  between  —  and  —  —  — 
b  n  n 

The  error,  therefore,  in  taking  either  of  these  values  for 

?  is  less  than  -.  But  by  increasing  n  indefinitely,  -  can  be 
on  n 

made  to  decrease  indefinitely,  and  to  become  less  than  any 
assigned  value,  however  small,  though  it  cannot  be  made 
absolutely  equal  to  zero. 

Hence,  the  ratio  of  two  incommensurable  quantities  cannot 
be  expressed  exactly  by  figures,  but  it  may  be  expressed  ap- 
proximately within  any  assigned  measure  of  precision. 

255,  The  ratio  of  two  incommensurable  quantities  is  called 
an  incommensurable  ratio  ;  and  is  a  fixed  value  toward  which 
its  successive  approximate  values  constantly  tend. 


94  PLANE   GEOMETRY.  —  BOOK   II. 

256.  THEOREM.    Two  incommensurable  ratios  are  equal  if, 

when  the  unit  of  measure  is  indefinitely  diminished,  their  ap- 
proximate values  constantly  remain  equal. 

Let  a:  b  and  a1 :  b*  be  two  incommensurable  ratios  whose  true 

values  lie   between   the   approximate   values  —  and  m       , 

n  n 

when  the  unit  of  measure  is  indefinitely  diminished.     Then 

they  cannot  differ  so  much  as  -• 

n 

Now  the  difference  (if  any)  between  the  fixed  values  a :  b 
and  a1 :  bf,  is  a,  fixed  value.  Let  d  denote  this  difference. 

Then  d<-. 

n 

But  if  d  has  any  value,  however  small,  -,  which  by  hypoth- 

n 

esis  can  be  indefinitely  diminished,  can  be  made  less  than  d. 

Therefore  d  cannot  have  any  value;  that  is,  d  =  Q,  and 
there  is  no  difference  between  the  ratios  a :  b  and  a' :  V ;  there- 
fore a  :  b  =  a!  :  b1. 

THE  THEORY  OF  LIMITS. 

257,  When  a  quantity  is  regarded  as  having  a  fixed  value 
throughout  the  same  discussion,  it  is  called  a  constant;  but 
when  it  is  regarded,  under  the  conditions  imposed  upon  it,  as 
having  different  successive  values,  it  is  called  a  variable. 

When  it  can  be  shown  that  the  value  of  a  variable,  measured 
at  a  series  of  definite  intervals,  can  by  continuing  the  series 
be  made  to  differ  from  a  given  constant  by  less  than  any 
assigned  quantity,  however  small,  but  cannot  be  made  abso- 
lutely equal  to  the  constant,  that  constant  is  called  the  limit 
of  the  variable,  and  the  variable  is  said  to  approach  indefi- 
nitely to  its  limit. 

If  the  variable  is  increasing,  its  limit  is  called  a  superior 
limit ;  if  decreasing,  an  inferior  limit. 


THEORY   OF   LIMITS.  95 

Suppose  a  point  to  move  from  A  toward  B,  under  the  con- 
ditions that  the  first  A  M  M<  M,  B 
second  it  shall  move 

one-half  the  distance  from  A  to  B,  that  is,  to  M ;  the  next 
second,  one-half  the  remaining  distance,  that  is,  to  M' ;  the 
next  second,  one-half  the  remaining  distance,  that  is,  to  M' ' ; 
and  so  on  indefinitely. 

Then  it  is  evident  that  the  moving  point  may  approach  as 
near  to  B  as  we  please,  but  will  never  arrive  at  B.  For,  how- 
ever near  it  may  be  to  B  at  any  instant,  the  next  second  it 
will  pass  over  one-half  the  interval  still  remaining ;  it  must, 
therefore,  approach  nearer  to  B,  since  half  the  interval  still 
remaining  is  some  distance,  but  will  not  reach  B,  since  half 
the  interval  still  remaining  is  not  the  whole  distance. 

Hence,  the  distance  from  A  to  the  moving  point  is  an  in- 
creasing variable,  which  indefinitely  approaches  the  constant 
AB  as  its  limit ;  and  the  distance  from  the  moving  point  to 
B  is  a  decreasing  variable,  which  indefinitely  approaches  the 
constant  zero  as  its  limit. 

If  the  length  of  AB  is  two  inches,  and  the  variable  is 
denoted  by  x,  and  the  difference  between  the  variable  and  its 
limit,  by  v : 

after  one  second,          x  =  1,  v  =  1 ; 

after  two  seconds,        x=  1  +  -J,  ^  =  -J- ; 

after  three  seconds,     #  =  1  + 1  +  i,  ^  —  i ; 

after  four  seconds,       #  =  l  +  2-  +  i  +  i>  v  =  •§• ; 
and  so  on  indefinitely. 

Now  the  sum  of  the  series  1  +  %  +  \  +  £,  etc.,  is  less  than 
2 ;  but  by  taking  a  great  number  of  terms,  the  sum  can  be 
made  to  differ  from  2  by  as  little  as  we  please.  Hence  2  is 
the  limit  of  the  sum  of  the  series,  when  the  number  of  the 
terms  is  increased  indefinitely ;  and  0  is  the  limit  of  the  dif- 
ference between  this  variable  sum  and  2. 


96 


PLANE   GEOMETRY.  —  BOOK   II. 


Consider  the  repetend  0.33333 ,  which  may  be  written 

TU  +  TITS  +  "nnrfr  +  nnnnr  + 

However  great  the  number  of  terms  of  this  series  we  take, 
the  sum  of  these  terms  will  be  less  than  -J- ;  but  the  more 
terms  we  take  the  nearer  does  the  sum  approach  -J-.  Hence 
the  sum  of  the  series,  as  the  number  of  terms  is  increased, 
approaches  indefinitely  the  constant  ^  as  a  limit. 

258.  In  the  right  triangle  AOB,  if  the  vertex  A  approaches 
indefinitely   the    base   BO,   the    angle   B  A 
diminishes,   and    approaches    zero    indefi- 
nitely ;  if  the  vertex  A  moves  away  from 

the  base  indefinitely,  the  angle  B  increases 
and  approaches  a  right  angle  indefinitely ; 
but  B  cannot  become  zero  or  a  right  angle, 
so  long  as  A  OB  is  a  triangle  ;  for  if  B  be- 
comes zero,  the  triangle  becomes  the  straight  line  BO,  and  if 
B  becomes  a  right  angle,  the  triangle  becomes  two  parallel 
lines  AO  &nd  AB  perpendicular  to  BO.  Hence  the  value  of 
B  must  lie  between  0°  and  90°  as  limits. 

259,  Again,  suppose  a  square  A  BOD  inscribed  in  a  circle, 
and  E,  F,  H,  _5Tthe  middle  points  of  the  arcs  subtended  by 
the  sides  of  the  square.     If  we  draw 

the  straight  lines  AE,  EB,  BF,  etc., 
we  shall  have  an  inscribed  polygon  of 
double  the  number  of  sides  of  the 
square. 

The  length  of  the  perimeter  of  this 
polygon,  represented  by  the  dotted 
lines,  is  greater  than  that  of  the 
square,  since  two  sides  replace  each 
side  of  the  square  and  form  with  it  a  triangle,  and  two  sides 
of  a  triangle  are  together  greater  than  the  third  side;  but  less 
than  the  length  of  the  circumference,  for  it  is  made  up  of 


THEORY   OF   LIMITS.  97 

straight  lines,  each  one  of  which  is  less  than  the  part  of  the 
circumference  between  its  extremities. 

By  continually  repeating  the  process  of  doubling  the  num- 
ber of  sides  of  each  resulting  inscribed  figure,  the  length  of 
the  perimeter  will  increase  with  the  increase  of  the  number 
of  sides ;  but  it  cannot  become  equal  to  the  length  of  the  cir- 
cumference, for  the  perimeter  will  continue  to  be  made  up  of 
straight  lines,  each  one  of  which  is  less  than  the  part  of  the 
circumference  between  its  extremities. 

The  length  of  the  circumference  is  therefore  the  limit  of  the 
length  of  the  perimeter  as  the  number  of  sides  of  the  inscribed 
figure  is  indefinitely  increased. 

260,  THEOREM.  If  two  variables  are  constantly  equal 
and  each  approaches  a  limit,  their  limits  are  equal. 

D 


Let  AM  and  AN  be  two  variables  which  are  con- 
stantly equal  and  which  approach  indefinitely  AB 
and  AC  respectively  as  limits. 

To  prove  AB  =  AC. 

Proof,   If  possible,  suppose  AB  >  AC,  and  take  AD  =  AC. 

Then  the  variable  A M  may  assume  values  between  AD  and 
AB,  while  tlae  variable  AN  must  always  be  less  than  AD. 
But  this  is  contrary  to  the  hypothesis  that  the  variables  should 

continue  equal. 

/.  AB  cannot  be  >  AC. 

In  the  same  way  it  may  be  proved  that  A C  cannot  be  >AB. 

.'.  AB  and  AC  are  two  values  neither  of  which  is  greater 
than  the  other. 

Hence  AB  =  A  C. 


98 


PLANE    GEOMETRY.  —  BOOK   II. 


MEASURE  OF  ANGLES. 

PROPOSITION  XVI.    THEOREM. 

261,  In  the  same  circle,  or  equal  circles,  two  angles 
at  the  centre  have  the  same  ratio  as  their  intercepted 
arcs. 


CASE  I.     When  the  arcs  are  commensurable. 

In  the  circles  whose  centres  are  C  and  D,  let  AGE  and 
EDF  be  the  angles,  AB  and  EF  the  intercepted  arcs. 

ove  £  ACB  _vn  AB 

(  ZEDF'^EF 

Proof,   Let  m  be  a  common  measure  of  AB  and  EF. 

Suppose  m  to  be  contained  in  AB  seven  times, 

and  in  EF  four  times. 

arc  AB  =  7 

arc  EF     4* 


Then 


(1) 


At  the  several  points  of  division  on  AB  and  EF  draw  radii. 
These  radii  will  divide  Z  ACB  into  seven  parts,  and 
EDF  into  four  parts,  equal  each  to  each,  §  229 

(in  the  same  O,  or  equal  ©,  equal  arcs  subtend  equal  4  at  the  centre). 


From  (1)  and  (2), 


'  Z  EDF    4 

Z.ACB  _ arc  AB 
'  arc  EF 


Ax.  1 


MEASURE   OF   ANGLES. 


CASE  II.    When  the  arcs  are  incommensurable. 
p 


In  the  equal  circles  ABP  and  A'B'P1  let  the  angles 
ACB  and  AfCf£f  intercept  the  incommensurable  arcs 
AB  and  A'B'. 

^  Z  ACB        arc  AB 

2o  prove  ,    Ai^-m  ~ TTTT/ 

Z  A'Q'J?      arc  A'B1 

Proof,  Divide  AB  into  any  number  of  equal  parts,  and 
apply  one  of  these  parts  as  a  unit  of  measure  to  A'B1  as  many 
times  as  it  will  be  contained  in  A'B1. 

Since  AB  and  A'Bf  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  A1  to  some  point,  as  D,  leav- 
ing a  remainder  DB'  less  than  one  of  these  parts. 

Draw  C'D. 


Since  AB  and  A'D  are  commensurable, 
Z  ACB  _arc4-g 


I. 


Z  A1  C'D     arc  A'D 

If  the  unit  of  measure  is  indefinitely  diminished,  these  ratios 
continue  equal,  and  approach  indefinitely  the  limiting  ratios 

ZAOB 
Z  A'C'B' 

Z  ACB 


arc  A'£' 


(If  two  variables  are  constantly  equal,  and  each  approaches  a  limit,  their 
limits  are  equal.) 


a.  E.  P 


100 


PLANE   GEOMETRY. — BOOK   II. 


262,  The  circumference,  like  the  angular  magnitude  about 
a  point,  is  divided  into  360  equal  parts,  called  degrees.     The 
arc-degree  is  subdivided  into  60  equal  parts,  called  minutes  ; 
and  the  minute  into  60  equal  parts,  called  seconds. 

Since  an  angle  at  the  centre  has  the  same  number  of  angle- 
degrees,  minutes,  and  seconds  as  the  intercepted  arc  has  of  arc- 
degrees,  minutes,  and  seconds,  we  say :  An  angle  at  the  centre 
is  measured  by  its  intercepted  arc ;  meaning,  An  angle  at  the 
centre  is  such  a  part  of  the  whole  angular  magnitude  about 
the  centre  as  its  intercepted  arc  is  of  the  whole  circumference. 

PKOPOSITION  XVII.     THEOREM. 

263,  An  inscribed  angle  is  measured  by  one-half 
the  are  intercepted  between  its  sides. 

B  B  B 


CASE  I.    When  one  side  of  the  angle  is  a  diameter. 
In   the   circle  PAS  (Fig.  1),  let  the  centre  C  be  in 
one  of  the  sides  of  the  inscribed  angle  B. 

To  prove         Z.  £  is  measured  by  ^  arc  PA. 
Proof.  Draw  CA. 

Radius  CA  =  radius  CB. 


§154 

But  ZPCA  =  ZB  +  ZA,  §145 

(the  exterior  Z.  of  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  <£). 


(being  opposite  equal  sides  of  the  A  CAB). 


But 


Z.  PCA  is  measured  by  PA, 
(the  £  at  the  centre  is  measured  by  the  intercepted  arc). 

.'.Z  B  is  measured  by  \  PA. 


§  262 


MEASURE   OP'  AJSTCiLIlS  101 

CASE  II.    When  the  centre  is  within  the  angle. 
In.   the  circle  BAE  (Fig.  2),   let  the  centre  C  fall 
within  the  angle  EBA. 
To  prove    Z  EBA  is  measured  by  \  arc  EA. 
Proof,  Draw  the  diameter  BCP. 

Z  PBA  is  measured  by  %  arc  PA,  Case  I. 

Z  PEE  is  measured  by  \  arc  PE,  Case  I. 

/.  Z  PBA  +  Z  PBE  is  measured  by  \  (arc  PA  +  arc  PE), 

or  Z  EBA  is  measured  by  ^  arc  EA. 
CASE  III.    When  the  centre  is  without  the  angle. 
In    the   circle  BFP  (Fig.   3),  let  the  centre  C  fall 
without  the  angle  ABF. 
To  prove    Z  ABF  is  measured  by  ^  arc  AF. 
Proof,  Draw  the  diameter  BCP. 

Z  PBF  is  measured  by  %  arc  PF,  Case  I. 

Z  PBA  is  measured  by  \  arc  PA.  Case  I. 

/.Z  PBF—/-  PBA  is  measured  by  \  (arc  PF—  arc  PA), 
or  Z  ^.-S^7  is  measured  by  %  arc  J..F.  Q.  E.  D. 


A 


FIG.  1.  FIG.  2.  FIG.  3. 


264,  COR.  1.  An  angle  inscribed  in  a  semicircle  is  a  right 
angle.     For  it  is  measured  by  one-half  a  semi-circumference. 

265,  Con.  2.  An  angle  inscribed  in  a  segment  greater  than  a 
semicircle  is  an  acute  angle.    For  it  is  measured  by  an  arc  less 
than  half  a  semi-circumference ;  as,  Z  CAD.     Fig.  2. 

266,  COR.  3.  An  angle  inscribed  in  a  segment  less  than  a 
semicircle  is  an  obtuse  angle.     For  it  is  measured  by  an  arc 
greater  than  half  a  semi-circumference  ;  as,  Z  CBD:    Fig.  2. 

267,  COR.  4.  All  angles  inscribed  in  the  sam.e  segment  are 
equal.     For  they  are  measured  by  half  the  same  arc.     Fig.  3. 


102  :PLANE   QEQMETRY. —  BOOK    II. 


PBOPOSITION  XVIII.     THEOREM. 

268,  An  angle  formed  by  two  chords  intersecting 
within  the  circumference  is  measured  ~by  one-half 
the  sum  of  the  intercepted  arcs. 


Let  the  angle  AOC  be  formed  by  the   chords  AB 
and  CD. 

To  prove    Z.  AOC  is  measured  by  ^  (AC+  BD). 
Proof,  Draw  AD. 

ZCOA  =  ^D  +  ZA,  §145 

(the  exterior  Z.  of  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A}. 
But  Z.  D  is  measured  by  -J-  arc  AC,  §  263 

and  Z  A  is  measured  by  ^  arc  BD, 

(an  inscribed  Z.  is  measured  by  J  the  intercepted  arc). 

c'.Z  CO  A  is  measured  by  £  (AC+  BD). 

Q.E.D. 


Ex.  83.  The  opposite  angles  of  an  inscribed  quadrilateral  are  sup- 
plements of  each  other. 

Ex.  84.  If  through  a  point  within  a  circle  two  perpendicular  chords 
are  drawn,  the  sum  of  the  opposite  arcs  which  they  intercept  is  equal  to 
a  semi-circumference. 

Ex.  85.  The  line  joining  the  centre  of  the  square  described  upon  the 
hypotenuse  of  a  rt.  A,  to  the  vertex  of  the  rt.  Z,  bisects  the  right  angle. 

HINT.  Describe  a  circle  upon  the  hypotenuse  as  diameter. 


MEASURE   OF   ANGLES. 


103 


PROPOSITION  XIX.    THEOREM. 

269,  An  angle  formed  by  a  tangent  and  a  chord  is 
measured  by  one-half  the  intercepted  arc. 


Let  MAH  be  the  angle  formed  by  the  tangent  MO 
and  chord  AH. 

To  prove    Z  MAH  is  measured  by  %  arc  A  EH. 
Proof.  Draw  the  diameter  ACF. 

Z  JO^isart.  Z,  §240 

(the  radius  drawn  to  a  tangent  at  the  point  of  contact  is  JL  to  it). 

Z  MAF  being  a  rt.  Z,  is  measured  by  \  the  semi-circum- 
ference AEF. 

But  Z  HAF  is  measured  by  J  arc  HF,  §263 

(an  inscribed  Z  is  measured  by  J  the  intercepted  arc). 

.'.  Z  MAF-/.  HAF  is  measured  by  \(AEF-  HF) ; 
or  Z  MAH  is  measured  by  |  AEH. 

Q.  E.  D. 

Ex.  86.  If  two  circles  touch  each  other  and  two  secants  are  drawn 
through  the  point  of  contact,  the  chords  joining  their  extremities  are 
parallel.  HINT.  Draw  the  common  tangent. 


104 


PLANE   GEOMETRY.  —  BOOK   II. 


PROPOSITION  XX.     THEOREM. 

270,  An  angle  formed  by  two  secants,  two  tangents, 
or  a  tangent  and  a  secant,  intersecting  without  the 
circumference,  is  measured  by  one-half  the  difference 
of  the  intercepted  arcs. 


FIG.  1. 

CASE  I.   Angle  formed  by  two  secants. 

Let  the  angle  0  (Fig.  1)  be  formed  by  the  two  se- 
cants OA  and  OB. 

To  prove      Z  0  is  measured  by  ^  (  AB  —  EG). 
Proof,  Draw  CB. 

ZAC£  =  ZO  +  ZB,  §145 

(the  exterior  Z  of  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A). 

By  taking  away  Z  B  from  both  sides, 


But 
and 


ZACBis  measured  by  £  AB, 

Z  B  is  measured  by  ^  CE, 
(an  inscribed  Z.  is  measured  by  J  the  intercepted  arc). 

.'.  Z  0  is  measured  by  \(AB—  CE). 


§  263 


MEASURE   OF   ANGLES.  105 

CASE  II.  Angle  formed  by  two  tangents. 

Let  the  angle  0  (Fig.  2)  be  formed  by  the  two  tan- 
gents OA  and  OB. 

To  prove  Z  0  is  measured  by  \  (A MB  —  ASB). 
Proof.  Draw  AB. 

Z  ABG=  Z  0  +  Z  OAB,  §  145 

(the  exterior  Z.  of  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A). 

By  taking  away  Z  OAB  from  both  sides, 
ZO  =  ZABC-ZOAB. 
But  Z  ABC  is  measured  by  \  AMB,  §  269 

and  Z  OAB  is  measured  by  %  ASB, 

(an  Z.  formed  by  a  tangent  and  a  chord  is  measured  by  J  the  intercepted  arc). 

.'.  Z  0  is  measured  by  £  (.4 JO—  .4/25). 

CASE  III.    Angle  formed  by  a  tangent  and  a  secant 

Let  the  angle  0  (Fig.  3)  be  formed  by  the  tangent 
OB  and  the  secant  OA. 

To  prove  Z  0  is  measured  by  %  (ADS  —  CJES). 
Proof,  Draw  OS. 

Z.ACS=ZO  +  ZCSO,  §145 

(the  exterior  /.  of  a  A  is  equal  to  the  sum  of  the  two  opposite  interior  A). 

By  taking  away  Z  080  from  both  sides, 
ZO-Z  ACS-  ZCSO. 

But  Z  ACS  IB  measured  by  £  ADS,  §  263 

(being  an  inscribed  Z), 

and  Z  CSO  is  measured  by  %CE8,  §  269 

(being  an  Z  formed  by  a  tangent  and  a  chord). 

.'.  Z  0  is  measured  by  ±(ADS—  CES). 

Q.E.  D. 


106  PLANE   GEOMETRY.  —  BOOK    II. 

PROBLEMS  OF  CONSTRUCTION. 
PROPOSITION   XXI.      PROBLEM. 

271.  At  a  given  point  in  a  straight  line,  to  erect  a 
perpendicular  to  that  line. 

\ 

" 


,} 

\J 


HOB  E^. 

FIG.  1.  FIG.  2. 

I.  Let  0  be  the  given  point  in  AC.  (Fig.  1). 
To  erect  a  J_  to  the  line  AC  at  the  point  0. 

Construction,  From  0  as  a  centre,  with  any  radius  OB, 
describe  an  arc  intersecting  AC  in  two  points  If  said  B. 

From  H  and  B  as  centres,  with  equal  radii  greater  than 
OB,  describe  two  arcs  intersecting  at  R.  Join  OR. 

Then  the  line  OR  is  the  J_  required. 

Proof,  Since  0  and  R  are  two  points  at  equal  distances  from 
.ZTand  B,  they  determine  the  position  of  a  perpendicular  to 
the  line  HB  at  its  middle  point  0.  §  123 

II.  When  the  given  point  is  at  the  end  of  the  line. 
Let  B  be  the  given  point.  (Fig.  2). 

To  erect  a  JL  to  the  line  AB  at  B. 

Construction,  Take  any  point  C  without  AB ;  and  from  C 
as  a  centre,  with  the  distance  CB  as  a  radius,  describe  an  arc 
intersecting  AB  at  E. 

Draw  EC,  and  prolong  it  to  meet  the  arc  again  at  D. 

Join  BD,  and  BD  is  the  J.  required. 

Proof.   The  Z.  B  is  inscribed  in  a  semicircle,  and  is  therefore 

a  right  angle.  §  264 

Hence  BD  is  _L  to  AB.  a  E.  F< 


PROBLEMS.  107 


PBOPOSITION  XXII.     PROBLEM. 

272,   Prom  a  point  without  a  straight  line,  to  let 
fail  a  perpendicular  upon  that  line. 


H  \T      M 


/ 
B 


Let  AB  be  the  given  straight  line,  and  C  the  given 
point  without  the  line. 

To  let  fall  a  J-  to  the  line  AS  from  the  point  C. 

Construction,   From  0  as  a  centre,  with  a  radius  sufficiently 
great,  describe  an  arc  cutting  AB  in  two  points,  .ZTand  K. 

From  jETand  K  &§  centres,  with  equal  radii  greater  than  \  HK, 
describe  two  arcs  intersecting  at  0. 

Draw  CO, 

and  produce  it  to  meet  AB  at  M. 
CM  is  the  JL  required. 

Proof,  Since  C  and  0  are  two  points  equidistant  from  H  and 
K,  they  determine  a  i.  to  HK  &i  its  middle  point.          §  123 

Q.E.F. 

NOTE.    Given  lines  of  the  figures  are  full  lines,   resulting  lines  are 
long-dotted,  and  auxiliary  lines  are  short-dotted. 


108  PLANE   GEOMETRY.  —  BOOK   II. 

PROPOSITION  XXIII.    PROBLEM. 
273,    To  bisect  a  given  straight  line. 


Let  AB  be  the  given  straight  line. 

To  bisect  the  line  AB. 

Construction,    From  A  and  B  as  centres,  with  equal  radii 
greater  than  ^  AB,  describe  arcs  intersecting  at  0  and  E. 

Join  CE. 
Then  the  line  CE  bisects  AB. 

Proof,    C  and  E  are  two  points  equidistant  from  A  and  B. 
Hence  they  determine  a  J.  to  the  middle  point  of  AB.    §  123 

Q.E.  F. 


Ex.  87.   To  find  in  a  given  line  a  point  X  which  shall  be  equidis- 
tant from  two  given  points. 

Ex.  88.   To  find  a  point  X  which  shall  be  equidistant  from  two 
given  points  and  at  a  given  distance  from  a  third  given  point. 

Ex.  89.  To  find  a  point  X  which  shall  be  at  given  distances  from 
two  given  points. 

Ex.  90.   To  find  a  point  X  which  shall  be  equidistant  from  threo 
given  points. 


PROBLEMS. 


109 


PROPOSITION  XXIV.   PROBLEM. 
274,  To  bisect  a  given  arc. 


Let  ACS  be  the  given  arc. 

To  bisect  the  arc  ACB. 

Construction,  Draw  the  chord  AB. 

From  A  and  B  as  centres,  with  equal  radii  greater  than 
£  A  B,  describe  arcs  intersecting  at  D  and  E, 

Draw  DE. 
DE  bisects  the  arc  ACB. 

Proof,  Since  D  and  E  are  two  points  equidistant  from  A 
and  B,  they  determine  a  J.  erected  at  the  middle  of  chord 
AB.  §  123 

And  a  J_  erected  at  the  middle  of  a  chord  passes  through 
the  centre  of  the  O,  and  bisects  the  arc  of  the  chord.  §  234 

Q.E.F. 

Ex.  91.  To  construct  a  circle  having  a  given  radius  and  passing 
through  two  given  points. 

Ex.  92.  To  construct  a  circle  having  its  centre  in  a  given  line  and 
passing  through  two  given  points. 


110  PLANE    GEOMETRY. BOOK    II. 

PROPOSITION  XXV.     PROBLEM. 
275,  To  bisect  a  given  angle. 


Let  AEB  be  the  given  angle. 

To  bisect  Z.  AEB. 

Construction.  From  E  as  a  centre,  with  any  radius,  as  EA, 
describe  an  arc  cutting  the  sides  of  the  Z  E  at  A  and  B. 

From  A  and  B  as  centres,  with  equal  radii  greater  than 
one-half  the  distance  from  A  to  J9,  describe  two  arcs  inter- 
secting at  C. 

Join  EC,  AC,  and  EC. 

EC  bisects  the  Z  E. 
Proof,    In  the  A  AEC  and  BEG 

AE  =  BE,  and  AC=  BQ,  Cons. 

and  EC--=-  EC.  Iden. 

.'.&AEC--=&BEC,  §160 

(having  three  sides  equal  each  to  each). 

.-.  Z  AEC=£BEC. 

Q.  E.  F. 

Ex.  93.   To  divide  a  right  angle  into  three  equal  parts. 
Ex.  94.    To  construct  an  equilateral  triangle,  having  given  one  side. 
Ex.  95.    To  find  a  point  X  which  shall  be  equidistant  from  two  given 
points  and  also  equidistant  from  two  given  intersecting  lines. 


PEOBLEMS.  Ill 


PROPOSITION   XXVI.     PROBLEM. 

276,  At  a  given  point  in  a  given  straight  line,  to 
construct  an  angle  equal  to  a  given  angle. 


c< 

E^ 

Let  C  be  the  given  point  in  the  given  line  CM,  and 
A  the  given  angle. 

To  construct  an  Z  at  0  equal  to  the  Z.  A. 

Construction,   From  A  as  a  centre,  with  any  radius,  as  AE, 
describe  an  arc  cutting  the  sides  of  the  Z  A  at  E  and  F. 

From  (7  as  a  centre,  with  a  radius  equal  to  AE, 

describe  an  arc  cutting  CM  at  H. 

From  H  as  a  centre,  with  a  radius  equal  to  the  distance  EF, 
describe  an  arc  intersecting  the  arc  HG  at  ra. 
Draw  Cm,  and  HCm  is  the  required  angle. 
Proof,  The  chords  EF&nd  Hm  are  equal.  Cons. 

/.  arc  EF=  arc  Hm,  §  230 
(in  equal  ©  equal  chords  subtend  equal  arcs). 

.-.Z.C=Z.A,  §229 

(in  equal  ©  equal  arcs  subtend  equal  A  at  the  centre).  Q.  E.  F. 


Ex.  96.  In  a  triangle  ABC,  draw  DE  parallel  to  the  base  BC,  cut- 
ting  the  sides  of  the  triangle   in  D  and  E,  so  that  DE  shall   equal 


Ex.  97.  If  an  interior  point  0  of  a  triangle  ABC  is  joined  to  the  ver- 
tices B  and  (7,  the  angle  BOG  is  greater  than  the  angle  BAC  of  the 
triangle. 


112  PLANE   GEOMETRY.  —  BOOK   II. 


PROPOSITION   XXVII.     PROBLEM. 

277,  Two  angles  of  a  triangle  being  given,  to  find 
the  third  angle. 


E  c_ 

E — 


I 

Let  A  and  B  be  the  two  given  angles  of  a  triangle. 
To  find  the  third  Z  of  the  A. 

Construction,   Take  any  straight  line,  as  EF,  and  at  any 
point,  as  Ht 

construct  Z  a  equal  to  Z  At  §  276 

and  Z  b  equal  to  Z  B. 
Then  Z  c  is  the  Z  required. 

Proof.   Since  the  sum  of  the  three  A  of  a  A  =  2  rt.  A,  §  138 
and  the  sum  of  the  three  A  a,  b,  and  c,  =  2  rt.  A\        §  92 
and  since  two  A  of  the  A  are  equal  to  the  A  a  and  b, 
the  third  Z  of  the  A  will  be  equal  to  the  Z  c.  Ax.  3. 


Q.  E.  F. 

Ex.  98.   In  a  triangle  ABC,  given  angles  A  and  B,  equal  respectively 
to  37°  13'  32"  and  41°  17'  56".     Find  the  value  of  angle  C. 


PROBLEMS.  H3 


PROPOSITION  XXVIII.  PROBLEM. 

278,   Through  a  given  point,  to  draw  a  straight  line 
parallel  to  a  given  straight  line. 


D 

Let  AB  "be  the  given  line,  and  C  the  given  point. 

To  draw  through  the  point  0  a  line  parallel  to  the  line  AB. 
Construction,   Draw  DCE,  making  the  Z  EDB. 

At  the  point  C  construct  Z  ECF=  Z  EDB.      §  276 
Then  the  line  FCH  is  II  to  AB. 

Proof.  Z  ECF=  Z  EDB.  Cons. 

.\ffFis  II  to  AB,  §108 

(when  two  straight  lines,  lying  in  the  same  plane,  are  cut  by  a  third  straight 
line,  if  the  ext.-int.  A  are  equal,  the  lines  are  parallel). 

Q.E.F. 


Ex.  99.  To  find  a  point  X  equidistant  from  two  given  points  and 
also  equidistant  from  two  given  parallel  lines. 

Ex.  100.   To  find  a  point  X  equidistant  from  two  given  intersecting 
lines  and  also  equidistant  from  two  given  parallels. 


114  PLANE   GEOMETRY. BOOK   II. 


PROPOSITION  XXIX.    PROBLEM. 

279,    To  divide  a  given  straight  line  into   equal 
parts. 

A^ — -. 7 TB 


w 


Let  AB  be  the  given  straight  line. 

To  divide  AB  into  equal  parts. 
Construction,        From  A  draw  the  line  AO. 

Take  any  convenient  length,  and  apply  it  to  AO  as  many 
times  as  the  line  AB  is  to  be  divided  into  parts. 

From  the  last  point  thus  found  on  AO,  as  C,  draw  CB. 

Through  the  several  points  of  division  on  AO  draw  lines 
II  to  CB,  and  these  lines  divide  AB  into  equal  parts. 

Proof,  Since  AC\&  divided  into  equal  parts,  AB  is  also,  §  187 

(if  three  or  more  \\s  intercept  equal  parts  on  any  transversal^  they  intercept 
equal  parts  on  every  transversal). 

Q.  E.  F. 


Ex.  101.   To  divide  a  line  into  four  equal  parts  by  two  different 
methods. 

Ex.  102.   To  find  a'  point  X  in  one  side  of  a  given  triangle  and  equi- 
distant from  the  other  two  sides. 

Ex.  103.  Through  a  given  point  to  draw  a  line  which  shall  make 
equal  angles  with  the  two  sides  of  a  given  angle. 


PKOBLEMS.  115 


PROPOSITION   XXX.     PROBLEM, 

280,   Two  sides  and  the  included  angle  of  a  trian- 
gle being  given,  to  construct  the  triangle. 

D 

s 

X 

\ 

y 


/ 
*>/ 


A/- — k L 

;         B 


Let  the  two  sides  of  the  triangle  "be  b  and  c,  and  the 
included  angle  A. 


To  construct  a  A  having  two  sides  equal  to  b  and  c  respec- 
tively, and  the  included  Z  =Z  A. 

Construction,   Take  AB  equal  to  the  side  c. 

At  A,  the  extremity  of  AB,  construct  an  angle  equal  to  the 
given  Z  A.  §  276 

On  AD  take  AC  equal  to  b. 

Draw  CB. 
Then  A  ACB  is  the  A  required. 

Q.E.F. 


Ex.  104.   To  construct  an  angle  of  45°. 

Ex.  105.  To  find  a  point  X  which  shall  be  equidistant  from  two 
given  intersecting  lines  and  at  a  given  distance  from  a  given  point. 

Ex.  106.  To  draw  through  two  sides  of  a  triangle  a  line  ||  to  the 
third  side  so  that  the  part  intercepted  between  the  sides  shall  have  a 
given  length. 


116  PLANE   GEOMETRY. — BOOK    II. 


PKOPOSITION  XXXI.     PROBLEM. 

281,   A  side  and  two  angles  of  a  triangle  being 
given,  to  construct  the  triangle. 


Let  c  be  the  given  side,  A  and  B  the  given  angles. 

To  construct  the  triangle. 

Construction,  Take  EC  equal  to  c. 

At  the  point  E  construct  the  Z  CEH  equal  to  Z  A.     §  276 

At  the  point  C  construct  the  Z  ECK  equal  to  Z  B. 

Let  the  sides  J^ZTand  OK  intersect  at  0. 
Then  A  COE  is  the  A  required. 

Q.  E.  F. 

REMARK.   If  one  of  the  given  angles  is  opposite  to  the  given  side, 
find  the  third  angle  by  $  277,  and  proceed  as  above. 

Discussion,  The  problem  is  impossible  when  the  two  given 
angles  are  together  equal  to  or  greater  than  two  right  angles. 


Ex.  107.  To  construct  an  angle  of  150°. 

Ex.  108.  A  straight  railway  passes  two  miles  from  a  town.  A  place 
is  four  miles  from  the  town  and  one  mile  from  the  railway.  To  find  by 
construction  how  many  places  answer  this  description. 

Ex.  109.  If  in  a  circle  two  equal  chords  intersect,  the  segments  of  one 
chord  are  equal  to  the  segments  of  the  other,  each  to  each. 

Ex.  110.  AB  is  any  chord  and  A  C  is  tangent  to  a  circle  at  A,  ODE  a 
line  cutting  the  circumference  in  D  and  E  and  parallel  to  AB\  show 
that  the  triangles  AGD  and  EAB  are  mutually  equiangular. 


PROBLEMS.  117 


PROPOSITION  XXXII.     PROBLEM. 

282,   The  three  sides  of  a  triangle  being  given*  to 
construct  the  triangle. 


/  v 


Let  the  three  sides  be  m,  71,  and  o. 

To  construct  the  triangle. 

Construction,  Draw  AB  equal  to  o. 

From  A  as  a  centre,  with  a  radius  equal  to  n,  describe  an 
arc  ; 

and  from  B  as  a  centre,  with  a  radius  equal  to  m,  describe 
an  arc  intersecting  the  former  arc  at  C. 

Draw  CA  and  CB. 
Then  A  CAB  is  the  A  required. 

Q.E.F. 

Discussion,  The  problem  is  impossible  when  one  side  is  equal 
to  or  greater  than  the  sum  of  the  other  two. 

Ex.  111.  The  base,  the  altitude,  and  an  angle  at  the  base,  of  a  tri- 
angle being  given,  to  construct  the  triangle. 

Ex.  112.  Show  that  the  bisectors  of  the  angles  contained  by  the  oppo- 
site sides  (produced)  of  an  inscribed  quadrilateral  intersect  at  right  angles. 

Ex.  113.  Given  two  perpendiculars,  AB  and  CD,  intersecting  in  0,  and 
a  straight  line  intersecting  these  perpendiculars  in  E  and  F\  to  construct 
a  square,  one  of  whose  angles  shall  coincide  with  one  of  the  right  angles 
at  0,  and  the  vertex  of  the  opposite  angle  of  the  square  shall  lie  in  EF. 
(Two  solutions.) 


118  PLANE   GEOMETRY.  —  BOOK   II. 


PROPOSITION  XXXIII.    PROBLEM. 

283,   Two  sides  of  a  triangle  and  the  angle  opposite 
one  of  them  being  given,  to  construct  the  triangle. 


/      \« 

a     \ 


../ 


-E 


— a  ^          (;/  £ 

CASE  I.    T/*  ^Ae  side  opposite  the  given  angle  is  less  than  the 
other  given  side. 

Let  b  be  greater  than  a,  and  A  the  given  angle. 

To  construct  the  triangle. 

Construction,   Construct  Z  DAE  —  to  the  given  Z  A.  §  276 

On  AD  take  AB  =  b. 

From  B  as  a  centre,  with  a  radius  equal  to  a, 
describe  an  arc  intersecting  the  line  AE  at  (7  and  (7'. 
Draw  JBO  arid  -5(7'. 

Then  both  the  A  ^4 £<?  and  ABC1  ,D 

fulfil  the  conditions,  and   hence  we  w  x 

have    two    constructions.      This    is 
called  the  ambiguous  case.  x^ 

Discussion,   If  the  side  a  is  equal          / 
to  the  JL  BIT,  the  arc  described  from 
B  will  touch  J.^7,  and  there  will  be 
but  one  construction,  the  right  tri- 
angle ABH.  B/''' 

If  the  given  side  a  is  less  than  the  /'  \a 

J.  from  B,  the  arc  described  from  B  /     \.- 

will  not  intersect  or  touch  AE,  and  ^ 

hence  the  problem  is  impossible, 


THE   CIRCLE.  119 

If  the  Z  A  is  right  or  obtuse,  the  prablem  is  impossible  ;  for  the 
side  opposite  a  right  or  obtuse  angle  is  the  greatest  side.  §  159 

CASE  II.  If  a  is  equal  to  b. 

If  the  Z  A  is  acute,  and  a  —  b,  the  arc  described  from  B  as 
a  centre,  and  with  a  radius  equal  to  a,  will 
cut  the  line  AE  at  the  points  A   and  C.  ^/ 

There  is   therefore   but   one  solution  :    the  &X  v? 

'•        X  N    > 

isosceles  A  ABC.  —  ~A^.  -  .^c  E 

Discussion,   If  the  Z  A  is  right  or  obtuse, 
the  problem  is  impossible  ;  for  equal  sides  of  a  A  have  equal 
A  opposite  them,  and  a  A  cannot  have  two  right  A  or  two 
obtuse  A. 

CASE  III.  If  a  is  greater  than  b. 

If  the  given  Z  A  is  acute,  the  arc  described  from  B  will  cut 
the  line  ED  on  opposite  sides  of  A,  at  C  and  C1.  The  A  ABQ 
answers  the  required  conditions,  but  the 


A  ABC1  does  not,  for  it  does  not  contain      .     a- 


the  acute  Z  A.    There  is  then  only  one  E  $( ^ 

solution  ;  namely,  the  A  ABC.  ^ -*' 

If  the  Z.  A  is  right,  the  arc  described 
from  B   cuts   the   line  ED  on  opposite  /\ 

sides  of  A,  and  we  have  two  equal  right  */'   |bv? 

A  which  fulfil  the  required  conditions.         1 


If  the  Z  J.  is  obtuse,  the  arc  described 
from  B  cuts  the  line   ED   on   opposite  \  « 

sides  of  A,  at  the  points  C  and  C".     The  a/'  ^N^ 

A  ABC  answers  the  required  conditions,   E    \/'         \\/  D 
but  the  A  ABC'  does  not,  for  it  does 
not  contain  the  obtuse  Z  A.     There  is  then  only  one  solu- 
tion ;  namely,  the  A  ABC. 


120  PLANE    GEOMETRY.  —  BOOK   II. 


PROPOSITION  XXXIV.  PROBLEM. 

284,   Two  sides  and  an  included  angle  of  a  paral- 
lelogram being  given,  to  construct  the  parallelogram. 


/ 

/ 


Let  m  and  o  be  the  two  sides,  and  C  the  included 
angle. 

To  construct  a  parallelogram. 
Construction,  Draw  AB  equal  to  o. 

At  A  construct  the  Z  A  equal  to  Z.  C,  §  276 

and  take  AIT  equal  to  m. 
From  H&s  a  centre,  with  a  radius  equal  to  o,  describe  an  arc. 

From  B  as  a  centre,  with  a  radius  equal  to  m, 
describe  an  arc,  intersecting  the  former  arc  at  E. 

Draw  .EZTand  EB. 
The  quadrilateral  ABEHis  the  O  required. 

Proof,  AB  =  HE,  Cons. 

AH=  BE.  Cons. 

/.  the  figure  ABEHis  a  O,  §  183 

(having  its  opposite  sides  equal). 

Q.E.  F. 


PROBLEMS.  121 


PROPOSITION  XXXV.     PROBLEM. 

285,    To  circumscribe  a  circle  about  a  given  tri- 
angle.  


Let  ABC  be  the  given  triangle. 

.     To  circumscribe  a  circle  about  ABO. 

Construction,  Bisect  AB  and  BO.  §  273 

At  the  points  of  bisection  erect  Js.  §  271 

Since  BC  is  not  the  prolongation  of  AB,  these  Js  will  in- 
tersect at  some  point  0. 

From  0,  with  a  radius  equal  to  OB,  describe  a  circle. 
O  ABC  is  the  0  required. 

Proof.    The  point  0  is  equidistant  from  A  and  B,  - 

and  also  is  equidistant  from  B  and  (7,  §  122 

(every  point  in  the  _L  erected  at  the  middle  of  a  straight  line  is  equidistant 
from  the  extremities  of  that  line). 

.'.  the  point  0  is  equidistant  from  A,  B,  and  C, 

and  a  O  described  from  0  as  a  centre,  with  a  radius  equal  to 
OB,  will  pass  through  the  vertices  A,  B,  and  O.  Q.E.F. 

286,  SCHOLIUM.  The  same  construction  serves  to  describe  a 
circumference  which  shall  pass  through  the  three  points  not 
in  the  same  straight  line  ;  also  to  find  the  centre  of  a  given 
circle  or  of  a  given  arc, 


122  PLANE    GEOMETRY.  —  BOOK   II. 


PROPOSITION   XXXVI.      PROBLEM. 

287,   Through  a  given  point,  to  draw  a  tangent  to  a 
given  circle. 


CASE  I.  When  the  given  point  is  on  the  circle. 

Let  C  "be  the  given  point  on  the  circle. 

To  draw  a  tangent  to  the  circle  at  C. 

Construction,   From  the  centre  0  draw  the  radius  0(7. 

Through  C  draw  A M  _L  to  OC.  §  271 

Then  AM  is  the  tangent  required. 

Proof,    A  straight  line  A-  to  a  radius  at  its  extremity  is  tan- 
gent to  the  circle.  §  239 

CASE  II.  When  the  given  point  is  without  the  circle. 
Let  O  be  the  centre  of  the  given  circle,  E  the  given 
point  -without  the  circle. 

To  draw  a  tangent  to  the  given  circle  from  the  point  E. 
Construction,  Join  OE. 

On  OE  as  a  diameter,  describe  a  circumference  intersecting 
the  given  circumference  at  the  points  J^and  -£T. 

Draw  OM  and  EM. 
Then  EM  is  the  tangent  required. 

Proof,  Z  OME  is  a  right  angle,  §  264 

(being  inscribed  in  a  semicircle). 

.'.  EM  is  tangent  to  the  circle  at  M.  §  239 

In  like  manner,  we  may  prove  HE  tangent  to  the  given  O. 

Q.  E.  F. 


PROBLEMS. 


123 


PROPOSITION   XXXVII.      PROBLEM. 
288,   To  inscribe  a  circle  in  a  given  triangle. 


Let  ABC  be  the  given  triangle. 

To  inscribe  a  circle  in  the  A  ABC. 

Construction,  Bisect  A  A  and  C.  §  275 

From  E,  the  intersection  of  these  bisectors, 

draw  EH  J_  to  the  line  AC.  §  272 

From  E,  with  radius  EH,  describe  the  O  KMH. 

The  O  KHMis  the  O  required. 

Proof,  Since  E  is  in  the  bisector  of  the  Z  A,  it  is  equidis- 
tant from  the  sides  AB  and  AC\  and  since  E  is  in  the  bisector 
of  the  Z  (7,  it  is  equidistant  from,  the  sides  AC  and  BC,  §  162 
(every  point  in  the  bisector  of  an  Z  is  equidistant  from  the  sides  of  the  Z). 
.'.  a  O  described  from  E  as  centre,  with  a  radius  equal  to 
EH,  will  touch  the  sides  of  the  A  and  be  inscribed  in  it. 


Q.  E.  F. 


289,  SCHOLIUM.  The  intersec- 
tions of  the  bisectors  of  exterior 
angles  of  a  triangle,  formed  by 
producing  the  sides  of  the  tri- 
angle, are  the  centres  of  three 
circles,  each  of  which  will  touch 
one  side  of  the  triangle,  and  the 
two  other  sides  produced.  These 
three  circles  are  called  escribed 
circles. 


124  PLANE   GEOMETRY. — BOOK   II. 


PROPOSITION  XXXVIII.    PROBLEM. 

290,    Upon  a  given  straight  line,  to  describe  a  seg- 
ment of  a  circle  which  shall  contain  a  given  angle. 


Let  AB  be  the  given  line,  and  M  the  given  angle. 

To  describe  a  segment  upon  AB  which  shall  contain  Z  M. 

Construction,  Construct  Z  ABE  equal  to  Z  M.  §  276 

Bisect  the  line  AB  by  the  _L  FO.  §  273 

From  the  point  B  draw  BO  _L  to  EB.  §  271 

From  0,  the  point  of  intersection  of  FO  and  BO,  as  a  cen- 
tre, with  a  radius  equal  to  OB,  describe  a  circumference. 
The  segment  AKB  is  the  segment  required. 

Proof,    The  point  0  is  equidistant  from  A  and  B,          §  122 
(every  point  in  a  J_  erected  at  the  middle  of  a  straight  line  is  equidistant 
from  the  extremities  of  that  line). 

.'.  the  circumference  will  pass  through  A. 
But  BE  is  _L  to  OB.  Cons. 

.-.  BE  is  tangent  to  the  O,  §  239 

(a  straight  line  JL  to  a  radius  at  its  extremity  is  tangent  to  the  O). 

/.  Z  ABE  is  measured  by  }  arc  AB,  §  269 

(being  an  /.formed  by  a  tangent  and  a  chord). 

An  Z  inscribed   in   the   segment  AKB   is   measured   by 

%AB.  §  263 

.'.  segment  AKB  contains  Z  M.  Ax.  1 


GL  E.  F. 


PROBLEMS.  125 


PROPOSITION  XXXIX.    PKOBLEM. 

291,  To  find  the  ratio  of  two  commensurable  straight 
lines. 


-B 


K 


F 

Let  AB  and  CD  be  two  straight  lines. 

To  find  the  ratio  of  AB  and  CD. 

Apply  CD.  to  AB  as  many  timers  possible. 

Suppose  twice,  with  a  remainder  EE. 
Then  apply  EE  to  CD  as  many  times  as  possible. 

Suppose  three  times,  with  a  remainder  FD. 
Then  apply  FD  to  EE  as  many  times  as  possible. 

Suppose  once,  with  a  remainder  HE. 
Then  apply  HE  to  FD  as  many  times  as  possible. 

Suppose  once,  with  a  remainder  KD. 

Then  apply  KD  to  HE  as  many  times  as  possible. 

Suppose  KD  is  contained  just  twice  in  HE. 

The  measure  of  each  line,  referred  to  KD  as  a  unit,  will 
then  be  as  follows  : 


=   ZKD-, 

EE  =  FD  +  HE  =  5  KD  ; 
CD  =  3  EE  +  FD  =  18  KD  ; 
+  EB= 


CD      18  KD  ' 
41  = 

CD          18  Q.E.F. 


...  the  ratio          =  — 


126  PLANE    GEOMETRY. BOOK    II. 

THEOREMS. 

114.  The  shortest  line  and  the  longest  line  which  can  be  drawn  from 
a  given  exterior  point  to  a  given  circumference  pass  through  the  centre. 

115.  If  through  a  point  within  a  circle  a  diameter  and  a  chord  _1_  to 
the  diameter  are  drawn,  the  chord  is  the  shortest  cord  that  can  be  drawn 
through  the  given  point. 

116.  In  the  same  circle,  or  in  equal  circles,  if  two  arcs  are  each 
greater  than  a  semi-circumference,  the  greater  arc  subtends  the  less 
chord,  and  conversely. 

117.   If  ABC  is  an  inscribed  equilateral  triangle,  and  P  is  any  point 
in  the  arc  BC,  then  PA  =  P£  +  PC. 

HINT.   On  PA  take  PM  equal  to  PB,  and  join  BM. 
%  118.  In  what  kinds  of  parallelograms  can  a  circle  be  inscribed  ? 
Prove  your  answer. 

119.  The  radius  of  the  circle  inscribed  in  an  equilateral  triangle  is 
equal  to  one- third  of  the  altitude  of  the  triangle. 

120.  A  circle  can  be  circumscribed  about  a  rectangle. 

121.  A  circle  can  be  circumscribed  about  an  isosceles  trapezoid. 

r^  122.  The  tangents  drawn  through  the  vertices  of  an  inscribed  rec- 
tangle enclose  a  rhombus. 

-f^  123.  The  diameter  of  the  circle  inscribed  in  a  rt.  A  is  equal  to  the 
difference  between  the  sum  of  the  legs  and  the  hypotenuse. 

124.  From  a  point  A  without  a  circle,  a  straight  line  AOB  is  drawn 
through  the  centre,  and  also  a  secant  ACD}  so  that  the  part  AC  without 
the  circle  is  equal  to  the  radius.     Prove  that  Z.  DAB  equals  one-third 
the  Z.  DOB. 

125.  All  chords  of  a  circle  which  touch  an  interior  concentric  circle 
are  equal,  and  are  bisected  at  the  points  of  contact. 

126.  If  two  circles  intersect,   and  a  secant   is  drawn  through  each 
point  of  intersection,  the  chords  which  join  the  extremities  of  the  secants 
are  parallel.      HINT.     By  drawing   the  common  chord,   two  inscribed 
quadrilaterals  are  obtained. 

127.  If  an  equilateral  triangle  is  inscribed  in  a  circle,  the  distance  of 
each  side  from  the  centre  of  the  circle  is  equal  to  half  the  radius. 

128.  Through   one   of  the   points   of  intersection   of  two   circles   a 
diameter  of  each  circle  is  drawn.     Prove  that  the  straight  line  joining 
the  ends  of  the  diameters  passes  through  the  other  point  of  intersection. 


EXERCISES.  127 

129.  A  circle  touches  two  sides  of  an  angle  £ACa,t  B,  C\  through  any 
point  D  in  the  arc  EC  a  tangent  is  drawn,  meeting  AB  at  E  &ud  AQ 
at  F.    Prove  (i.)  that  the  perimeter  of  the  triangle  AEF  is  constant  for 
all  positions  of  D  in  BC\  (ii.)  that  the  angle  EOF  is  also  constant. 

Loci. 

130.  Find  the  locus  of  a  point  at  three  inches  from  a  given  point. 

131.  Find  the  locus  of  a  point  at  a  given  distance  from  a  given 
circumference. 

132.  Prove  that  the  locus  of  the  vertex  of  a  right  triangle,  having  a 
given  hypotenuse  as  base,  is  the  circumference  described  upon  the  given 
hypotenuse  as  diameter. 

133.  Prove  that  the  locus  of  the  vertex  of  a  triangle,  having  a  given 
base  and  a  given  angle  at  the  vertex,  is  the  arc  which  forms  with  the 
base  a  segment  capable  of  containing  the  given  angle. 

134.  Find  the  locus  of  the  middle  points  of  all  chords  of  a  given 
length  that  can  be  drawn  in  a  given  circle. 

135.  Find  the  locus  of  the  middle  points  of  all  chords  that  can  be 
drawn  through  a  given  point  A  in  a  given  circumference. 

?v    136.  Find  the  locus  of  the  middle  points  of  all  straight  lines  that  can 

pe  drawn  from  a  given  exterior  point  A  to  a  given  circumference. 

137.   A  straight  line  moves  so  that  it  remains  parallel  to  a  given  line, 

and  touches  at  one  end  a  given  circumference.     Find  the  locus  of  the 

other  end. 
,    1 138.   A  straight  rod  moves  so  that  its   ends  constantly  touch  two 

fi&ed  rods  which  are  _L  to  each  other.   Find  the  locus  of  its  middle  point. 

139.  In  a  given  circle  let  AOB  be  a  diameter,  OC  any  radius,  CD 
the  perpendicular  from  C  to  AB.    Upon  OC  take  OM=CD.    Find  the 
locus  of  the  point  If  as  OC  turns  about  0. 

CONSTRUCTION  OF  POLYGONS. 

To  construct  an  equilateral  A,  having  given : 

140.  The  perimeter.          141.   The  radius  of  the  circumscribed  circle, 
142.  The  altitude.  143.   The  radius  of  the  inscribed  circle. 

To  construct  an  isosceles  triangle,  having  given: 
144,  The  angle  at  the  vertex  and  the  base. 


128  PLANE   GEOMETRY.  —  BOOK   II. 

145.  The  angle  at  the  vertex  and  the  altitude. 

146.  The  base  and  the  radius  of  the  circumscribed  circle. 

147.  The  base  and  the  radius  of  the  inscribed  circle. 

148.  The  perimeter  and  the  alti- 
tude. 

HINTS.   Let  ABO  be  the  A  re-  C 

quired,  and  EF  the  given  perimeter. 
The  altitude  CD  passes  through  the  ^ 

middle  of  EF,  and   the  &  AEC, 
BFG  are  isosceles.  E  3    5 

To  construct  a  right  triangle,  having  given : 

149.  The  hypotenuse  and  one  leg. 

150.  The  hypotenuse  and  the  altitude  upon  the  hypotenuse. 

151.  One  leg  and  the  altitude  upon  the  hypotenuse  as  base. 

.  The  median  and  the  altitude  drawn  from  the  vertex  of  the  rt.  Z. 

153.  The  radius  of  the  inscribed  circle  and  one  leg. 

154.  The  radius  of  the  inscribed  circle  and  an  acute  angle. 

155.  An  acute  angle  and  the  sum  of  the  legs. 

156.  An  acute  angle  and  the  difference  of  the  legs. 

To  construct  a  triangle,  having  given : 

>  157.  The  base,  the  altitude,  and  the  Z  at  the  vertex. 

Xl58.  The  base,  the  corresponding  median,  and  the  Z  at  the  vertex. 

159.  The  perimeter  and  the  angles.  \^\ 

-  160.  One  side,  an  adjacent  Z,  and  the  sum  of  the  other  sides. 

161.  One  side,  an  adjacent  Z,  and  the  difference  of  the  other  sides. 

**  162.  The  sum  of  two  sides  and  the  angles. 

163.  One  side,  an  adjacent  Z,  and  radius  of  circumscribed  O. 

164.  The  angles  and  the  radius  of  the  circumscribed  O. 

165.  The  angles  and  the  radius  of  the  inscribed  O. 

166.  An  angle,  the  bisector,  and  the  altitude  drawn  from  the  vertex. 

167.  Two  sides  and  the  median  corresponding  to  the  other  side. 

168.  The  three  medians. 

To  construct  a  square,  having  given : 

169.  The  diagonal.        170^The  sum  of  the  diagonal  and  one  side. 


EXERCISES.  129 

To  construct  a  rectangle,  having  given: 

171.  One  side  and  the  Z  formed  by  the  diagonals. 

172.  The  perimeter  and  the  diagonal. 

•      173.   The  perimeter  and  the  Z*of  the  diagonals. 

174.  The  difference  of  the  two  adjacent  sides  and   the  Z  of  the 
diagonals. 

To  construct  a  rhombus,  having  given : 

175.  The  two  diagonals. 

176.  One  side  and  the  radius  of  the  inscribed  circle. 
'~~^177.   One  angle  and  the  radius  of  the  inscribed  circle. 

178.  One  angle  and  one  of  the  diagonals. 

To  construct  a  rhomboid,  having  given: 

179.  One  side  and  the  two  diagonals. 

180.  The  diagonals  and  the  Z  formed  by  them. 

181.  One  side,  one  Z,  and  one  diagonal. 

182.  The  base,  the  altitude,  and  one  angle. 

To  construct  an  isosceles  trapezoid,  having  given: 

183.  The  bases  and  one  angle.        184.  The  bases  and  the  altitude. 

185.  The  bases  and  the  diagonal. 

186.  The  bases  and  th   radius  of  the  circumscribed  circle. 

To  construct  a  trapezoid,  having  given : 

187.  The  four  sides.        188.   The  two  bases  and  the  two  diagonals. 
1S9.  The  bases,  one  diagonal,  and  the  Z  formed  by  the  diagonals. 

CONSTRUCTION  OF  CIRCLES. 

Find  the  locus  of  the  centre  of  a  circle : 

190.  Which  has  a  given  radius  r  and  passes  through  a  given  point  P. 

191.  Which  has  a  given  radius  r  and  touches  a  given  straight  line  AB. 

192.  Which  passes  through  two  given  points  Pand  Q. 

193.  Which  touches  a  given  straight  line  AB  at  a  given  point  P. 

194.  Which  touches  each  of  two  given  parallels. 

195.  Which  touches  each  of  two  given  intersecting  lines. 


130  PLANE   GEOMETRY. BOOK   II. 

To  construct  a  circle  which  has  the  radius  r  and  which  also : 

196.  Touches  each  of  two  intersecting  lines  AB  and  CD. 

197.  Touches  a  given  line  AB  and  a  given  circle  K. 

198.  Passes  through  a  given  point  P  and  touches  a  given  line  AB. 

199.  Passes  through  a  given  point  P  and  touches  a  given  circle  K. 

To  construct  a  circle  which  shall : 

200.  Touch  two  given  parallels  and  pass  through  a  given  point  P. 

201.  Touch  three  given  lines  two  of  which  are  parallel. 

202.  Touch  a  given  line  AB  at  P  and  pass  through  a  given  point  Q. 

203.  Touch  a  given  circle  at  P  and  pass  through  a  given  point  Q. 

204.  Touch  two  given  lines  and  touch  one  of  them  at  a  given  point  P. 

205.  Touch  a  given  line  and  touch  a  given  circle  at  a  point  P. 

206.  Touch  a  given  line  AB  at  P  and  also  touch  a  given  circle. 

207.  To  inscribe  a  circle  in  a  given  sector. 

%  '208.  To  construct  within  a  given  circle  three  equal  circles,  so  that 
each  shall  touch  the  other  two  and  also  the  given  circle. 

/./209.  To  describe  circles  about  the  vertices  of  a  given  triangle  as 
centres,  so  that  each  shall  touch  the  two  others. 


CONSTRUCTION  OF  STRAIGHT  LINES. 

/  210.  To  draw  a  common  tangent  to  two  given  circles. 

211.   To  bisect  the  angle  formed  by  two  lines,  without  producing  the 
lines  to  their  point  of  intersection. 

t  212.   To  draw  a  line  through  a  given  point,  so  that  it  shall  form  with 
.the  sides  of  a  given  angle  an  isosceles  triangle. 

213.  Given  a  point  P  between  the  sides  of  an  angle  BAG.    To  draw 
through  P  a  line  terminated  by  the  sides  of  the  angle  and  bisected  at  P. 
"J214.  Given  two  points  P,  Q,  and  a  line  AB-  -to  draw  lines  from  P 
and  Q  which  shall  meet  on  AB  and  make  equal  angles  with  AB. 

HINT.   Make  use  of  the  point  which  forms  with  P  a  pair  of  points 
symmetrical  with  respect  to  AB. 

7  215.  To  find  the  shortest  path  from  Pto  Q  which  shall  touch  a  line  AB. 
^  ^  216.  To  draw  a  tangent  to  a  given  circle,  so  that  it  shall  be  parallel 
to  a  given  straight  line. 


BOOK  III. 

PROPORTIONAL    LINES    AND    SIMILAR 
POLYGONS. 


THE  THEORY  OF  PROPORTION. 

292,  A  proportion  is  an  expression  of  equality  between  two 
equal  ratios. 

A  proportion  may  be  expressed  in  any  one  of  the  follow- 
ing forms : 

—  =  —  ;     a  :  b >•  =  c :  d\     a: b : : c : d; 
b      a 

and  is  read,  "  the  ratio  of  a  to  b  equals  the  ratio  of  c  to  d" 

293,  The  terms  of  a  proportion  are  the  four  quantities  com- 
pared ;  the  first  and  third  terms  are  called  the  antecedents,  the 
second  and  fourth  terms,  the  consequents ;  the  first  and  fourth 
terms  are  called  the  extremes,  the  second  and  third  terms,  the 
means. 

294,  In  the  proportion  a  :  b  =  c  :  d,  d  is  a  fourth  propor- 
tional to  a,  5,  and  c. 

In  the  proportion  a  :  b  =  b  :  c,  c  is  a  third  proportional  to 
a  and  b. 

In  the  proportion  a:b  —  b:ct  b  is  a  mean  proportional 
between  a  and  c. 


132  PLANE   GEOMETRY.  —  BOOK   III. 

PROPOSITION  I. 

295,  In  every  proportion  the  product  of  the  extremes 
is  equal  to  the  product  of  the  means. 

Let  a:b  =  c:d. 
To  prove  ad  =  be. 

Now  ?  =  £, 

o     d 

whence,  by  multiplying  both  sides  by  bdt 

ad  =  be.  a  Et  a 

PROPOSITION  II. 

296,  A  mean  proportional  between  two  Quantities 
is  equal  to  the  square  root  of  their  product. 

In  the  proportion  a  :  b  =  b  :  c, 

b*  =  ac,  §  295 

(the  product  of  the  extremes  is  equal  to  the  product  of  the  means). 

Whence,  extracting  the  square  root, 

b  =  -Vac.  Q.E.D. 

PROPOSITION  III. 

297,  If  the  product  of  two  quantities  is  equal  to  the 
product  of  two  others,  either  two  may  be  made  the 
extremes  of  a  proportion  in  which  the  other  two  are 

made  the  means. 

Let  ad=  be. 

To  prove  a:b  =  c:  d. 

Divide  both  members  of  the  given  equation  by  bd. 

M 

or,  a :  b  =  c  :  d.  o.  E.  o. 


THEORY   OF   PROPORTION.  133 


PROPOSITION  IV. 

298,  If  four  quantities  of  the  same  kind  are  in  pro- 
portion, tJiey  will  be  in  proportion  by  alternation; 
that  is,  the  first  term  will  be  to  the  third  as  the  sec- 
ond to  the  fourth. 

Let  a:  6  —  c:d. 

To  prove  a:  c  —  b  :d. 

Now  f?=r£. 

Multiply  each  member  of  the  equation  by  -. 

c 

Then  2  =  *, 

c      d 

or,  a:c  =  b:d. 

Q.E.D. 

PROPOSITION  V. 

299,  If  four  quantities  are  in  proportion,  they  will 
be  in  proportion  fry  inversion ;  that  is,  the  second  term 
will  be  to  the  first  as  the  fourth  to  the  third. 

Let  a:  b  =  c:  d. 

To  prove  b  :  a  =  d :  c . 

Now  be  =  ad.  §  295 

Divide  each  member  of  the  equation  by  ac. 

Then  £  =  £ 

a     c 

or,  b :  a  —  d :  c. 

a  to, 


134  PLANE   GEOMETRY.  —  BOOK   III. 

PROPOSITION  VI. 

300,  If  four  quantities  are  in  proportion,  they  will 
be  in  proportion  by  composition  ;  that  is,  the  sum  of 
the  first  two  terms  will  be  to  the  second  term  as  the 
sum  of  the  last  two  terms  to  the  fourth  term. 

Let  a:b=c:d. 
To  prove  a  +  b  :b  =  c  +  d:d. 

-NT  a     c 

Now  -  =  -- 

b      a 

Add  1  to  each  member  of  the  equation. 

Then  £+1  =  ^+1; 

b  a 

,,,.  a  +  b      c  +  d 

that  is,  —  ;  —  =  —  4—' 

b  d 

or,  a  +  b  :  b  =  c  +  d  :  d. 

In  like  manner,      a  +  b  :a=c  +  d:  c. 

Q.E.  D. 

PROPOSITION  VII. 

301,  If  four  quantities  are  in  proportion^  they  will 
be  in  proportion  by  division  \  that  is,  the  difference 
of  the  first  two  terms  will  be  to  the  second  term  as 
the  difference  of  the  last  two  terms  to  the  fourth 

term. 

Let  a-.b  =  c:d. 

To  prove  a  —  b:b  =  c  —  d:d. 

AT  a      c 

Now  I  =  3' 

b      d 

Subtract  1  from  each  member  of  the  equation. 
Then 

that  is,  «ZL=2r, 

6  d 

or,  a  —  b:b  —  c  —  d:d. 

In  like  manner,     a  —  b:a  =  c—  -die.  ae.a 


Then  £-  !=§-!; 

b  d 


c 

C  — 


THEORY   OF    PROPORTION.  135 

PROPOSITION  VIII. 

302,  In  any  proportion  the  terms  are  in  proportion 
by  composition  and  division;  that  is,  the  sum  of  the 
first  two  terms  is  to  their  difference  as  the  sum  of 
the  last  two  terms  to  their  difference. 

Let  a:b  =  c:d. 

Then,  by  §  300, 
And,  by  §  301, 

By  division,  _, 

a  —  b      c  —  d 

or,  a  +  b  :  a  —  b—  c  +  d:  c  —  d. 

Q.  E.  D. 

PROPOSITION  IX. 

303,  In  a  series  of  equal  ratios,  the  sum  of  the  an- 
tecedents is  to  the  sum  of  the  consequents  as  any 
antecedent  is  to  its  consequent. 

Let  a:b  =  c:  d=e  :f=  g :  h. 
To  prove     a+c+e+g  :b-\-d+f+h  =  a  :b. 
Denote  each  ratio  by  r. 

mi  d         C          6         O 

r=i=d=jh 

Whence,      a  =  br,     c  =  dr,     e  =fr,    g  =  hr. 
Add  these  equations. 

Then  a  +  c-\-e-\- g  =  (b  + d -\-f-\-  K)r. 

Divide  by  (b  +  d+f+  K). 


Then  a     *      g     .y=        * 

b+d+f+h  b 

or,  a+c  +  e  +  g:b  +  d  +/+  h  —  a:b. 

Q.E.D. 


136  PLANE   GEOMETRY.  —  BOOK   III. 

PROPOSITION  X. 

304,  The  products  of  the  corresponding  terms  of 
two  or  more  proportions  are  in  proportion. 

Leta:b  =  c:d,    e:f=g:h,    k:l  =  min. 

To  prove  aek  :  bfl  =  cgm  :  dhn. 

TVT™  ace      a     k     m 

Now  l=-?    7      *'    7=  — 

b      d  f      h    I      n 

Whence,  by  multiplication, 

aek  __  cgm 
bfl  "dhn 
or,  aek  :  bfl  =  cgm  :  dhn. 

CLE.D. 

PROPOSITION  XI. 

305,  Like  powers,  or  like  roots,  of  the  terms  of  a 
proportion  are  in  proportion. 

Let  a:b  =  cid. 
To  prove  an:bn  =  cn:dn, 

III1 

and  an  :  fa  =  c»  :  d*. 

Now  2=4 

b      d 

By  raising  to  the  nth  power, 

^•=~n\  or  an:bn  =  cn:dn. 
o      a 

By  extracting  the  nth  root, 
i        i 


On          M 
U,n  (yn 

—  =  —  ;  or, 


1111 


Q.E.O. 

306,  Equimultiples  of  two  quantities  are  the  products  ob- 
tained by  multiplying  each  of  them  by  the  same  number. 
Thus,  ma  and  mb  are  equimultiples  of  a  and  b. 


THEORY   OF   PROPORTION.  137 


PROPOSITION  XII. 

307,  Equimultiples  of  two  quantities  are  in  the 
same  ratio  as  the  quantities  themselves. 

Let  a  and  b  be  any  two  quantities. 
To  prove  ma  :  mb  =  a:b. 

•vr  a     a 

Now  -=-• 

Multiply  both  terms  of  first  fraction  by  m. 

Then  ~  =  7* 

mb      b 

or,  ma  :  mb  =  a  :  b. 

Q.E.D. 

308i  SCHOLIUM.  In  the  treatment  of  proportion  it  is  as- 
sumed that  fractions  may  be  found  which  will  represent  the 
ratios.  It  is  evident  that  the  ratio  of  two  quantities  may  be 
represented  by  a  fraction  when  the  two  quantities  compared 
can  be  expressed  in  integers  in  terms  of  a  common  unit.  But 
when  there  is  no  unit  in  terms  of  which  both  quantities  can  be 
expressed  in  integers,  it  is  possible  to  find  a  fraction  that  will 
represent  the  ratio  to  any  required  degree  of  accuracy.  (See 
§§  251-256.) 

Hence,  in  speaking  of  the  product  of  two  quantities,  as  for 
instance,  the  product  of  two  lines,  we  mean  simply  the  product 
of  the  numbers  which  represent  them  when  referred  to  a  com- 
mon unit. 

An  interpretation  of  this  kind  must  be  given  to  the  product 
of  any  two  quantities  throughout  the  Geometry. 


138 


PLANE   GEOMETRY.  —  BOOK   III. 


PROPORTIONAL  LINES. 
PROPOSITION  I.    THEOREM. 

309,  If  a  line  is  drawn  through  two  sides  of  a  tri- 
angle parallel  to  the  third  side,  it  divides  those  sides 
proportio  nally. 

A  4 


FIG.  2. 


In  the  triangle  ABC  let  EF  be  drawn  parallel  to  BC. 
„  EB     FQ 

Toprove  AE=  AF 

CASE  I.  When  AE  and  EB  (Fig.  1)  are  commensurable. 
Find  a  common  measure  of  AE  and  EB,  as  BM. 
Suppose  BM  to  be  contained  in  BE  three  times, 
and  in  AE  four  times. 


The, 


(I) 


g=| 

At  the  several  points  of  division  on  BE  and  AE  draw 
straight  lines  II  to  BC. 

These  lines  will  divide  AC  mto  seven  equal  parts,  of  which 
EC  will  contain  three,  and  A.  vvill  contain  four,  §  187 

(if  parallels  intercept  equal  parts  on  any  transversal,  they  intercept  equal 
parts  on  every  transversal). 

.  F0      3  <X 


Compare  (1)  and  (2), 


EB_ 
AE' 


Ax.  1. 


PROPORTIONAL    LINES.  139 

CASE  II.  When  AE  and  EB  (Fig.  2)  are  incommensurable. 

Divide  AE  into  any  number  of  equal  parts,  and  apply  one 
of  these  parts  as  a  unit  of  measure  to  EB  as  many  times  as  it 
will  be  contained  in  EB. 

Since  AE  and  EB  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  E  to  a  point  K,  leaving  a 
remainder  KB  less  than  the  unit  of  measure. 
Draw  KH  II  to  BO. 

-Then 

Suppose  the  unit  of  measure  indefinitely  diminished,  the 


ratios  —  —  and  -  continue    equal;    and    approach    indefi- 
AE          AF 


nitely  the  limiting  ratios  —  —  and  —rji,  respectively. 


Therefore  :?f=TV  §260 


Q.  E.  D. 

310,  COR.  1.  One  side  of  a  triangle  is  to  either  part  cut  off 
by  a  straight  line  parallel  to  the  base  as  the  other  side  is  to  the 
corresponding  part. 

For  EB:AE=FC:  AF,  by  the  theorem. 

/.  EB  +  AE:*AE=  FC+  AF:  AF,  §  300 

or  AB:  AE=AC\  AF. 

311,  COR.  2.  If  two  lines  are  cut  by  any  number  of  parallels, 
the  corresponding  intercepts  are  proportional. 

•A.  (J 

Let  the  lines  be  AB  and  CD.  Ft\L     \G 

Draw  AN  II  to  CD,  cutting  the  Us  at  L,  M,  '      \        \ 

and^V.     Then  *    f     f 

AL^CG,  LM^GK,  MN^KD.    §187     /     \       \ 
By  the  theorem,  B     N       D 

AIT:  AM=  AF:  AL  =  FIT:  LM=  HB  :  MN. 

That  is,  AF  :  CG  =  FH:  GK=  HB  :  KD. 

If  the  two  lines  AB  and  CD  were  parallel,  the  correspond- 

ing intercepts  would  be  equal,  and  the  above  proportion  be  true. 


140  PLANE   GEOMETRY.  —  BOOK   III. 


PROPOSITION  II.     THEOREM. 
>- 

312,  If  a  straight  line  divides  two  sides  of  a  tri- 
angle proportionally,  it  is  parallel  to  the  third  side. 


In  the  triangle  ABC  let  EF  be  drawn  so  that 

AB^AC 
AE     AF 

To  prove  EF  \\toBO. 

Proof,  From  E  draw  EH  II  to  BG. 

Then  AB:  AE^AC:  AH,  §310 

(one  side  of  a  A  is  to  either  part  cut  off  by  a  line  II  to  the  base,  as  the  other 
side  is  to  the  corresponding  part). 


But  AB  :  AE  =  AC:  AF.  Hyp. 

The  last  two  proportions  have  the  first  three  terms  equal, 
each  to  each  ;  therefore  the  fourth  terms  are  equal  ;  that  is, 

AF=AH. 

/.  ^Fand  EH  coincide. 

But  EH  is  II  to  EC.  Cons. 

/.  EF,  which  coincides  with  EH,  is  ||  to  BC. 


PROPORTIONAL   LINES.  141 


PROPOSITION  III.    THEOREM. 

313,   The  bisector  of  an  angle  of  a  triangle  divides 
the  opposite  side  into  segments  proportional  to 
other  two  sides. 


AM  B 

Let  CM  bisect  the  angle  C  of  the  triangle  CAB. 

To  prove  MA  :  MB  =  CA  :  CB. 

Proof,   Draw  AE  II  to  MG  to  meet  BG  produced  at  E. 
Since  MG  is  II  to  AE  of  the  A  BAE,  we  have          §  309 
MA:MB=GE-.GB.  (1) 

Since  MG  is  II  to  AE, 

ZACM=ZCAE,  §104 

(being  alt  Ant.  Aof\\  lines) ; 

and  Z  J50M=  Z  GEA,  §  106 

(being  ext.-int.  A  of  II  lines). 

But  the  Z  A  CM=  Z  BCM.  Hyp. 

.-.  the  Z  GAE  =  Z  CEA.  Ax.  1 

.-.  GE=  CA,  §  156 

(if  two  A  of  a  A  are  e^waZ,  £Ae  opposite  sides  are  equal), 

Putting  CA  for  GE  in  (1),  we  have 

MA  :  JO  =  C4  :  GB. 

Q.E.  D. 


142  PLANE   GEOMETRY.  —  BOOK   III. 


PROPOSITION  IV.     THEOREM. 

314,  The  bisector  of  an  exterior  angle  of  a  triangle 
meets  the  opposite  side  produced  at  a  point  the  dis- 
tances of  which  from  the  extremities  of  this  side  are 
proportional  to  the  other  two  sides. 


A 

Let  CM'  bisect  the  exterior  angle  ACE  of  the  tri- 
angle CAB,  and  meet  BA  produced  at  M'. 

To  prove  M'A  :  M'B  =*  OA  :  CB. 

Proof,       Draw  AF  II  to  CM'  to  meet  BO  at  F. 
Since  AFis  II  to  CM1  of  the  A  BCM*  ',  we  have  §  309 

M'A:M'B=--CF'.CB.  (1) 

Since  AF  is  II  to  CM1, 

the  Z  M'CE  =  Z  AFC,  §  106 

(being  ext.-int.  Aof\\  lines)  \ 

and  the  Z  M'CA  =  Z  OAF,  §  104 

(being  alt.-ini.  A  of  II  lines). 

Since  CM1  bisects  the  Z  ECA, 


.'.  the  Z  AFC  =  Z  CAF.  Ax.  1 

.*.  CA  =  OF,  §  156 

(if  two  A  of  a  A  are  equal,  the  opposite  sides  are  equal). 

Putting  CA  for  CFin.  (1),  we  have 

M'A  :  M'£  -  CA  :  CB. 

Q.E.Q, 


PROPORTIONAL    LINES.  143 

315,  SCHOLIUM.  If  a  given  line  AB  is  divided  at  M,  a 
point  between  the  extremities  A  and  B,  it  is  said  to  be 
divided  internally  into  the  segments  MA  and  MB  ;  and  if  it 
is  divided  at  Mf,  a  point  in  the  prolongation  of  AB,  it  is  said 
to  be  divided  externally  into  the  segments  M1  A  and  M  'JB. 


-B 


A          M 

In  either  case  the  segments  are  the  distances  from  the  point 
of  division  to  the  extremities  of  the  line.  If  the  line  is  divided 
internally,  the  sum  of  the  segments  is  equal  to  the  line  ;  and 
if  the  line  is  divided  externally,  the  difference  of  the  segments 
is  equal  to  the  line. 

Suppose  it  is  required  to  divide  the  given  line  AB  inter- 
nally and  externally  in  the  same  ratio  ;  as,  for  example,  the 
ratio  of  the  two  numbers  3  and  5. 


M  '  A          M  B        y 

We  divide  AB  into  5  +  3,  or  8,  equal  parts,  and  take  8 
parts  from  A  ;  we  then  have  the  point  M,  such  that 

MA:M£  =  3:5.  (1) 

Secondly,  we  divide  AB  into  two  equal  parts,  and  lay  off 
on  the  prolongation  of  AB,  to  the  left  of  A,  three  of  these 
equal  parts  ;  we  then  have  the  point  M  f,  such  that 

M'A  :  M*B  -3:5.  (2) 

Comparing  (1)  and  (2), 

MA:MB  =  M'A  :  M  '£. 


316,  If  a  given  straight  line  is  divided  internally  and 
externally  into  segments  having  the  same  ratio,  the  line  is 
said  to  be  divided  harmonically. 


144  PLANE   GEOMETRY.  —  BOOK    III. 

317.  COR.  1.  The  bisectors  of  an  interior  angle  and  an  exte- 
rior angle  at  one  vertex  of  a  triangle  j 

divide    the   opposite  side  harmoni- ' 

catty.    For,  by  §§  313  and  314,  each 

bisector  divides  the  opposite  side 

into  segments  proportional  to  the 

o^her  two  sides  of  the  triangle.  M1          "A     M~ 

318.  COR.  2.  If  the  points  M  and  M'  divide  the  line  AB 
harmonically,  the  points  A  and  B  divide  the  line  MM1  har- 
monically. 

For,  if  MA  :  MB  -  M  'A  :  M1  B, 

by  alternation,      MA  :  M'A  =  MB  :  M'B.  §  298 

That  is,  the  ratio  of  the  distances  of  A  from  M  and  Mf  is 
equal  to  the  ratio  of  the  distances  of  B  from  M  and  M '. 

The  four  points  A,  B,  M,  and  Mf  are  called  harmonic 
points,  and  the  two  pairs,  A,  B,  and  M,  M\  are  called  con- 
jugate harmonic  points. 

SIMILAR  POLYGONS. 

319.  Similar  polygons  are  polygons  that  have  their  homol- 
ogous angles  equal,  and  their  homologous  sides  proportional. 

B 


ED  E'          D' 

Thus,  if  the  polygons  ABODE  and  A'B'C'D'E'  are  similar 
the  A  A,  B,  (7,  etc.,  are  equal  to  A  A1,  B1,  C1,  etc. 


-0.  -L>  .D  \J  CD 

a  d  7^= Wo<=7*n<'  etc' 

jfi  jD  -ft   \j  \j    l J 

320.   In  two  similar  polygons,  the  ratio  of  any  two  homol- 
ogous sides  is  called  the  ratio  of  similitude  of  the  polygons. 


SIMILAR  TRIANGLES.  145 

SIMILAR  TRIANGLES. 
PROPOSITION  V.     THEOREM. 

321,  Two  mutually  equiangular  triangles  are  sim- 
ilar. 

A 

A 


In  the  triangles'  ABG  and  A'B'C1  let  angles  A,  B,  C  be 
equal  to  angles  A',  Bf,  Cr  respectively. 

To  prove  A  ABC  and  A'B'C'  similar. 

Proof,         Apply  the  A  A'B'C'  to  the  A  ABO, 
so  that  Z.  A1  shall  coincide  with  /.  A. 
Then  the  A  A'B'C'  will  take  the  position  of  A  AEH. 
Now  /.  AJSff(s&me  as  Z  B1)  =  Z  B. 

.'.  JEHis  II  toJStf,  §108 

(when  two  straight  lines,  lying  in  the  same  plane,  are  cut  by  a  third  straight 
Line,  if  the  ext.-int.  4  are  equal  the  lines  are  parallel). 

.-.  AB  :  AE=  AC:  AH,  §  310 

or  AJ3:A'B'  =  AC:A'C'. 

In  like  manner,  by  applying  A  A'B'O'  to  A  ABC,  so  that 
Z  B1  shall  coincide  with  Z  B,  we  may  prove  that 

AB  :  A'£'  =  BO:  JB'0*. 
Therefore  the  two  A  are  similar.  §  319 

Q.E.D. 

322,  COR.  1.   Two  triangles  are  similar  if  two  angles  of  the 
one  are  equal  respectively  to  two  angles  of  the  other. 

323,  COR.  2.    Two  right  triangles  are  similar  if  an  acute 
angle  of  the  one  is  equal  to  an  acute  angle  of  the  other. 


146  PLANE   GEOMETRY.  —  BOOK   III. 


PROPOSITION  VI.     THEOREM. 

324,   If  two  triangles  have  their  sides  respectively 
proportional,  they  are  similar. 


In  the  triangles  ABC  and  A'B'C?  let 

AB  =  AC  =  BC 
A'B'     A'C'     B'C1' 

To  prove          A  ABC  and  A' B'C1  similar. 
Proof,         Take  AE  =  A'B',  and  AH=  A'C'. 

Draw  EH. 
Then  from  the  given  proportion, 

AB  =  AC 
AE     AH 

/.  EH  is  II  to  BO,  §  312 

(if  a  line  divide  two  sides  of  a  A  proportionally,  it  is  II  to  the  third  side). 

Hence  in  the  A  ABC  and.  AEH 

Z  ABO=  Z  AEH,  §  106 

and  Z.ACB  =  Z.AHE, 

(being  ext.-int.  Aofll  lines). 

.%  A  ABC&nd  AEH  are  similar,  §  322 

(two  &  are  similar  if  two  A  of  one  are  equal  respectively  to  two  A  of  the 

other). 

;.AB:  AE  =BC:EH; 
that  is,  A£:A'£'  =  £C:  EH. 


SIMILAR  TRIANGLES. 


147 


But  by  hypothesis, 


The  last  two  proportions  have  the  first  three  terms  equal, 
each  to  each  ;  therefore  the  fourth  terms  are  equal  ;  that  is, 

EH=  B'C'. 
Hence  in  the  A  AEHznd  A'  B'C', 

]£H=B'C',  AE=A'B',  w&AH=A'C'. 

/.  A  AEJET=  A  A'B'C',  §  160 

(having  three  sides  of  the  one  equal  respectively  to  three  sides  of  the  other). 

But  A  AEH  is  similar  to  A  ABC. 

.-.  A  A'B'C'  is  similar  to  A  ABO.  aE.D. 

325,  SCHOLIUM.  The  primary  idea  of  similarity  is  likeness 
of  form  ;  and  the  two  conditions  necessary  to  similarity  are  : 

I.  For  every  angle  in  one  of  the  figures  there  must  be  an 
equal  angle  in  the  other,  and 

II.  The  homologous  sides  must  be  in  proportion. 

In  the  case  of  triangles,  either  condition  involves  the  other, 
but  in  the  case  of  other  polygons,  it  does  not  follow  that  if  one 
condition  exist  the  other  does  also. 


Q 


Q' 


Thus  in  the  quadrilaterals  Q  and  Q',  the  homologous  sides 
are  proportional,  but  the  homologous  angles  are  not  equal. 

In  the  quadrilaterals  E  and  R1  the  homologous  angles  are 
equal,  but  the  sides  are  not  proportional. 


148  PLANE   GEOMETKY. —  BOOK   III. 


PBOPOSITION  VII.    THEOREM. 

326,  If  two  triangles  have  an  angle  of  the  one  equal 
to  an  angle  of  the  other,  and  the  including  sides  pro- 
portional, they  are  similar. 


In  the  triangles  ABC  and  A'B'C',  let  Z4  =  Z4',  and 

AB  =  AC 
A'B'~  A'C'' 

To  prove          A  ABC  and  A'£'C'  similar. 

Proof,   Apply  the  A  A'£'C'  to  the  A  ABC,  so  that  Z  A! 
shall  coincide  with  Z  A. 

Then  the  A  A'£'C'  will  take  the  position  of  A  AEH. 


That  is 

AE~  AH 

Therefore  the  line  EH  divides  the  sides  AB  and  .4  (7  pro- 
portionally ; 

.-.  ^^Tis  II  to  EC,  §  312 

(if  a  line  divide  two  sides  of  a  A  proportionally,  it  is  II  to  the  third  side). 

Hence  the  A  ABC  and  AEH  are  mutually  equiangular 
and  similar. 

.-.  A  A'£'C'  is  similar  to  A  AEQ. 

Q.E.  O. 


SIMILAR   TRIANGLES. 


149 


PROPOSITION  VIII.     THEOREM. 

327,  If  two  triangles  have  their  sides  respectively 
parallel,  or  respectively  perpendicular,  they  are  sim- 
ilar. 

A 


B' 


C' 


In  the  triangles  A'B'C'  and  ABC  let  A'B',  A'C',  B'a  be 
respectively  parallel,  or  respectively  perpendicular, 
to  AB,  AC,  BC. 

To  prove  .        A  A'H'C1  and  A  BC  similar. 

• 

Proof,  The  corresponding  A  are  either  equal  or  supplements 
of  each  other,  §§  112,  113 

(if  two  A  have  their  sides  II,  or  JL,  they  are  equal  or  supplementary). 
Hence  we  may  make  three  suppositions  : 

1st,  A  +  A'  =  2rt.A,    B  +  N  =  2  rt.  A, 
2d.  A  =  A',          £  +  £'  =  2rt.A, 

3d.  A  =--  A1,  B  =  ®,      :.  0=  C'.          §  140 

Since  the  sum  of  the  A  of  the  two  A  cannot  exceed  four 
right  angles,  the  third  supposition  only  is  admissible.  §  138 


/.  the  two  A  ABC  and  A'3'C9  are  similar, 
(two  mutually  equiangular  &  are  similar). 


§  321 

Q.E.  O. 


150 


PLANE   GEOMETRY.  —  BOOK    III. 


PROPOSITION  IX.     THEOREM. 

328,  The  homologous  altitudes  of  two  similar  tri- 
angles have  the  same  ratio  as  any  two  homologous 
sides. 


A  o 


A'    0 


In  the  two  similar  triangles  ABC  and  A'B'C',  let  the 
altitudes  be  CO  and  C'0f. 


To  prove 


CO        AC       AB 


C'O'     A'C'      A'ff 


Proof,    In  the  rt.  A  CO  A  and  C'O' A', 

Z  A  ==  Z  A1,  §  319 

(being  homologous  A  of  the  similar  &  ABC  and  A'B'C1). 

.-.  A  CO  A  and  C'O' A'  are  similar,  §  323 

(two  rt.  &  having  an  acute  Z  of  the  one  equal  to  an  acute  Z  of  the  other 
are  similar). 

CO        AC 


'  C'O'     A'C' 

In  the  similar  A  ABC  *&&.  A'£'C', 
AC       AB 


§319 


Therefore, 


A'C'     A'B' 

CO  =  AC  =  AB 
C'O1     A'C1     A'ff 


Q.  E.  O. 


SIMILAR   TRIANGLES.  151 


PROPOSITION  X.     THEOREM. 

> 

329,  Straight  lines  drawn  through  the  same  point 
intercept  proportional  segments  upon  two  parallels. 


ABC         D        E 

Let  the  two  parallels  AE  and  A'E'  cut  the  straight 
lines  OA,  OBt  OC,  OD,  and  OE. 

AB       BO       CD        DE 


To  prove 


A'ff     B1C*      C'D'      D'E' 


Proof,  Since  A'E1  is  II  to  AE,  the  pairs  of  A  OAB  and 
OA'-B1,  OBC  and  Oll'C',  etc.,  are  mutually  equiangular  and 
similar, 

_OB_      d  BO  _OB 

~  ~ 


(homologous  sides  of  similar  A  are  proportional). 

AB  =    EC 
"  A'B'     B'C' 

In  a  similar  way  it  may  be  shown  that 

BO       CD        ,  CD        DE 

and- 


Ax.  1 


_£'<?'      C'D'         C'D'     D'E' 

0.1*0, 

REMARK.   A  condensed  form  of  writing  the  above  is 

AB  =  (OB\  =  _BC_  =  fOO\  =  _CD_  =  / OD\  =  D^ 
^'j5'      \OB')      B'C'     \00'J      C'&     \ODf)     D'Ef 

where  a  parenthesis  about  a  ratio  signifies  that  this  ratio  is  used  to 
prove  the  equality  of  the  ratios  immediately  preceding  and  following  it. 


152  PLANE  GEOMETRY.  —  BOOK   III. 


PROPOSITION  XI.     THEOREM. 

330,  CONVERSELY  :  If  three  or  more  non-parallel 
straight  lines  intercept  proportional  segments  upon 
two  parallels,  they  pass  through  a  'common,  point. 


AC  E 

Let  AB,  CD,  EF,  cut  the  parallels  AE  and  BF  so  that 
AC:  BD=CE  :  DF. 

To  prove  that  AB,  CD,  EF prolonged  meet  in  a  point. 
Proof,   Prolong  AB  and  CD  until  they  meet  in  0. 
Join  OE. 

If  we  designate  by  F  the  point  where  OE  cuts  BF,  we 
shall  have  by  §  329, 

AC:BD=CE-.DF'. 
But  by  hypothesis 

AC:BD=CE:DF. 

These  proportions  have  the  first  three  terms  equal,  each  to 
each ;  therefore  the  fourth  terms  are  equal ;  that  is, 

DF'  =  DF. 

.'.  F1  coincides  with  F. 

.'.  EF  prolonged  passes  through  O. 

.'.  AB.  CD.  and  EF  prolonged  meet  in  the  point  0. 

Q.E.D. 


SIMILAR    POLYGONS. 


153 


SIMILAR  POLYGONS. 
PROPOSITION  XII.     THEOREM. 

331,  If  two  polygons  are  composed  of  the  same  num- 
ber of  triangles,  similar  each  to  each,  and  similarly 
placed,  the  polygons  are  similar. 
E 


B  G  B'  c' 

In  the  two  polygons  ABODE  and  A'B'C'D'E1,  let  the 
triangles  AEB,  BEC,  CED  be  similar  respectively  to 
the  triangles  A'E'B',  B'E'C',  C'EfD>. 

To  prove       ABODE  similar  to  A'B'C'D'E1. 

Proof,  Z  A  =  Z  A\  §  319 

(being  homologous  A  of  similar  &). 

Also,  Z  ABE  =  Z  A'B'E',  §  319 

and  Z  EEC  -  Z  E'B'C'. 


By  adding,  Z  ABC  =  Z  A'JB'C'. 

In  like  manner  we  may  prove  Z  BCD  =  Z  JB'C'D',  etc. 
Hence  the  two  polygons  are  mutually  equiangular. 
Now 

AE  ^  AB  =(  EB\      BC  _^(  EC\       CD  _    ED 
A'E'     AB1     \E'B'J     B'C1     \E'(?)     C'D'     E'D1' 

(the  homologous  sides  of  similar  A  are  proportional). 
Hence  the  homologous  sides  of  the  polygons  are  proportional. 

Therefore  the  polygons  are  similar,  §  319 

(having  their  homologous  A  equal,  and  their  homologous  sides  proportional). 

Q.  E.  0, 


154 


PLANE   GEOMETRY.  —  BOOK   III. 


PROPOSITION  XIII.     THEOREM. 

332,  If  two  polygons  are  similar,  they  are  composed 
of  the  same  number  of  triangles,  similar  each  to  each, 
and  similarly  placed. 
E 


B  C  B'  C 

Let  the  polygons  ABODE  and  A'B'C'D'E1  be  similar. 

From  two  homologous  vertices,  as  E  and  E\  draw  diagonals 
EBt  EC,  and  £'£',  E'C'. 

To  prove  A  EAB,  EEC,  ECD 

similar  respectively  to  A  E'A'B',  E'3'C',  E'C'D'. 

Proof,   In  the  A  EAB  and 


and 


(being  homologous  A  of  similar  polygons) ; 
AE       AB 


§319 
§319 
§326 


(being  homologous  sides  of  similar  polygons). 
.-.  A  EAB  and  E'AB*  are  similar, 

(having  an  /.  of  the  one  equal  to  an  Z  of  the  other,  and  the  including  sides 
proportional). 

Also,  Z  ABC=  /.  A'B'C',  (1) 

(being  homologous  A  of  similar  polygons). 

And  Z  ABE=Z.  AB'E\  (2) 

(being  homologous  A  of  similar  A). 
Subtract  (2)  from  (1), 

£EBC=/.EIB'C*.  Ax.  3 


SIMILAR   POLYGONS. 


155 


Now 


And 


EB        AB 


E'B'     A'B' 

(being  homologous  sides  of  similar  A). 

BO  =  AB 

B'C'     A'B'J 

(being  homologous  sides  of  similar  polygons). 

.    EB  _  BC 
"E'B'     B'C1' 

:.  A  EBCsmd  E'B'C'  are  similar, 


Ax.  1 


§326 


(having  an  /.  of  the  one  equal  to  an  Z  of  the  other,  and  the  including  sides 
proportional). 

In  like  manner  we  may  prove  A  ECD  and  E'C'D*  similar. 

Q.  E.  D. 

PROPOSITION  XIV.     THEOREM. 

333,  The  perimeters  of  two  similar  polygons  have 
the  same  ratio  as  any  two  homologous  sides. 
E 


B  C  Bf  C 

Let  the  two  similar  polygons  be  ABODE  and  A'&C'D'E', 
and  let  P  and  P1  represent  their  perimeters. 
To  prove  P:  P' =  AB  :  A'B'. 

AB  :  A'B'  =  BC:  B'C'  =  CD  :  C<D',  etc,        §  319 
(the  homologous  sides  of  similar  polygons  are  proportional). 

.-.  AB+BC,etc. :  A'B' +  B'C',  eic.^AB-.A'B1,  §  303 

(in  a  series  of  equal  ratios  the  sum  of  the  antecedents  is  to  the  sum  of  the 
consequents  as  any  antecedent  is  to  its  consequent). 

That  is,  P:P*  =  AB:  A'B'.  Q.  E  D. 


156  PLANE  GEOMETRY.  —  BOOK  III. 

NUMERICAL  PROPERTIES  OF  FIGURES. 
PROPOSITION  XV.     THEOREM. 

^     334,  If  in  a  right  triangle  a  perpendicular  is  drawn 
from  the  vertex  of  the  right  angle  to  the  hypotenuse : 

I.  The  perpendicular  is  a  mean  proportional  be- 
tween the  segments  of  the  hypotenuse. 

II.  Each  leg  of  the  right  triangle  is  a  mean  pro- 
portional between  the  hypotenuse  and  its  adjacent 
segment. 


F 

In  the  right  triangle  ABC,  let  BF  be  drawn  from  the 
vertex  of  the  right  angle  B,  perpendicular  to  AC. 

I.  To  prove  AF:  BF==  BF:  FC. 
Proof,   In  the  rt.  A  BAFmA.  BAG 

the  acute  Z  A  is  common. 
Hence  the  A  are  similar.  §  323 

In  the  rt.  A  BCF&ud  BOA 

the  acute  Z  (7  is  common. 
Hence  the  A  are  similar.  .  §  323 

Now  as  the  rt.  A  ABF and  CBFwc*  both  similar  to  ABO, 
they  are  similar  to  each  other. 

In  the  similar  A  ABF&U&  CBF, 

AF,  the  shortest  side  of  the  one, 
:  BF,  the  shortest  side  of  the  other, 
:  :  BF,  the  medium  side  of  the  one, 
:  FC,  the  medium  side  of  the  other. 

II.  To  prove          AC:  AB  =  AB  :  AF, 
and 


NUMERICAL    PROPERTIES   OF    FIGURES.  157 

In  the  similar  A  ^LBtf  and  ABF, 

AC,  the  longest  side  of  the  one, 
:  AB,  the  longest  side  of  the  other, 
:  :  AB,  the  shortest  side  of  the  one, 

:  AF,  the  shortest  side  of  the  other. 
Also  in  the  similar  A  ABC  and  FBC, 

AC,  the  longest  side  of  the  one, 
:  EC,  the  longest  side  of  the  other, 
:  :  EC,  the  medium  side  of  the  one, 

:  FC,  the  medium  side  of  the  other.  Q.  E.  o. 

335,    COR.  1.  The  squares  of  the  two  legs  of  a  right  triangle 
are  proportional  to  the  adjacent  segments  of  the  hypotenuse. 
The  proportions  in  II.  give,  by  §  295, 

Al?  =  ACxAF,  and  BG*  =  ACx  CF. 
By  dividing  one  by  the  other,  we  have 
ACxAF_AF 


^O2     ACx  CF      CF 

336,  COR.  2.    The  squares  of  the  hypotenuse  and  either  leg 
are  proportional  to  the  hypotenuse  and  the  adjacent  segment. 

1^  _/_L  (_/    -/j-O  X.  ^JL  O    .^j.  O 

~£&  =  ACxAF~~~AF 

337,  COR.  3.    An  angle  inscribed  in  a  semicircle  is  a  right 
angle  (§  264).     Therefore, 

I.  The  perpendicular  from  any  point  in 
the  circumference  to  the  diameter  of  a  circle 

is  a  mean  proportional  between  the  segments  //_ 
of  the  diameter. 

II.  The  chord  drawn  from  the  point  to  either  extremity  of  the 
diameter  is  a  mean  proportional  between  the  diameter  and  the 
adjacent  segment. 

REMAEK.  The  pairs  of  corresponding  sides  in  similar  triangles  may  be 
called  longest,  shortest,  medium,  to  enable  the  beginner  to  see  quickly 
these  pairs  ;  but  he  must  not  forget  that  two  sides  are  homologous,  not 
because  they  appear  to  be  the  longest  or  the  shortest  sides,  but  because 
they  lie  opposite  corresponding  equal  angles. 


158 


PLANE    GEOMETRY. 


BOOK    III. 


PROPOSITION  XVI.     THEOREM. 

338,  The  sum  of  the  squares  of  the  two  legs  of  a  right 
triangle  is  equal  to  the  square  of  the  hypotenuse. 


Let  ABC  be  a  right  triangle  with  its  right  angle  at  B. 

To  prove  AB*  +  BC2  =  AC*. 

Proof,  Draw  BF J.  to  AC. 

Then  AB2  =  ACxAF  §334 

and  BC2  =  ACx  CF 

By  adding,  AB2  +  BC2  =  A C(AF+  CF)  =  AC\        a  E.  D. 

339,  COR.    The  square  of  either  leg  of  a  right  triangle  is  equal 
to  the  difference  of  the  squares  of  the  hypotenuse  and  the  other  leg. 

340,  SCHOLIUM.    The  ratio  of  the  diagonal  of  a  D 
square  to  the  side  is  the  incommensurable  num- 
ber A/2.     For  if  A  C  is  the  diagonal  of  the  square 

A  BCD,  then 

AC2  =  AB2  +  BC\  or  AC2  =  2 AB\ 


/ 

Divide  by  AB  ,  we  have  4      =  2,  or  ^x  =, 


Since  the  square  root  of  2  is  incommensurable,  the  diagonal 
and  side  of  a  square  are  two  incommensurable  lines. 

341,  The  projection  of  a  line  CD  upon  a  straight  line  AB  is 
that  part  of  the  line  AB  comprised 
between  the  perpendiculars  CP  and 
DR  let  fall  from  the  extremities  of 
CD.  Thus,  PR  is  the  projection  of 
CD  upon  AB. 


A- 


NUMERICAL   PROPERTIES   OF   FIGURES.  159 


PROPOSITION  XVII.     THEOREM. 

342,  In  any  triangle,  the  square  of  the  side  opposite 
an  acute  angle  is  equal  to  the  sum  of  the  squares  of 
the  other  two  sides  diminished  ~by  twice  the  product 
of  one  of  those  sides  and  the  projection  of  the  other 
upon  that  side. 

A 


Let  C  be  an  acute  angle  of  the  triangle  ABC,  and 
DC  the  projection  of  AC  upon  BC. 

To  prove      A32  --=  BC2  +  AC*  -  2  BCx  DO. 
Proof,    If  D  fall  upon  the  base  (Fig.  1), 

DB  =  BC-  DC] 
If  D  fall  upon  the  base  produced  (Fig.  2), 

DB  =  DC-~BC. 
In  either  case, 

DB2  ^Btf  +  DC2 -2BCx  DO. 
Add  AD1  to  both  sides  of  this  equality,  and  we  have 
AD2  +  DB2  =  BC2  +  AD2+DC2-2BCx  DC. 
But  AD2  +  DB2  =  AB\  §  338 

and  AD*  +  DC2  =  AC\ 

(the  sum  of  the  squares  of  the  two  legs  of  a  rt.  A  is  equal  to  the  square 
of  the  hypotenuse). 

Put  AB  and  AC   for  their  equals  in  the  above  equality, 
BC*  +  AC2-  2£Cx  DC. 

Q.  E.  D. 


160  PLANE  GEOMETRY.  —  BOOK   III. 


PKOPOSITION  XVIII.    THEOREM. 

343,  In  any  obtuse  triangle,  the  square  of  the  side 
opposite  the  obtuse  angle  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides  increased  by  twice  the 
product  of  one  of  those  sides  and  the  prqjectipn  of 
the  other  upon  that  side. 
A 


c 

Let  C  be  the  obtuse  angle  of  the  triangle  ABC,  and 
CD  be  the  projection  of  AC  upon  BC  produced. 

To  prove      AB*  =  BC*  +  AC2  +  2BCX  DG. 
Proof,  DB^BC+DC. 

Squaring,   DB*  =  BC*  +  DC*  +  2  BCx  DO. 
Add  AD  to  both  sides,  and  we  have 

AD*  +  DB*  =  BC*+AD*+DC*+2BCxDC. 
But  AD*  +  DB*  =  AB\  §  338 


and  XD2  +  DC*  =  AC\ 

(the  sum  of  the  squares  of  the  two  legs  of  a  rt.  A  is  equal  to  the  square 
of  the  hypotenuse). 

Put  AJ?  and  AC*  for  their  equals  in  the  above  equality, 
AW  =  BC*  +  AC*  +  2  BCx  DC. 

Q.E.D. 

NOTE.   The  last  three  theorems  enable  us  to  compute  the  lengths  of 
the  altitudes  if  the  lengths  of  the  three  sides  of  a  triangle  are  known. 


NUMERICAL   PROPERTIES   OF    FIGURES.          .      161 

•^     PROPOSITION  XIX.     THEOREM. 

344,  I.  The  sum  of  the  squares  of  two  sides  of  a  tri- 
angle is  equal  to  twice  the  square  of  half  the  third 
side  increased  by  twice  the  square  of  the  median  upon 
that  side. 

II.  The  difference  of  the  squares  of  two  sides  of  a 
triangle  is  equal  to  twice  the  product  of  the  third 
side  by  the  projection  of  the  median  upon  that  side. 


u 

In  the  triangle  ABG  let  AM  "be  the  median,  and  MD 
the  projection  of  AM  upon  the  side  BG.  Also  let  AB 
be  greater  than  AC. 

To  prove      I.  Aff  +  AC?  =  2  BM*  +  2  AM\ 

II.  JO?-  AC2  =  2BCx  MD. 

Proof,  Since  AB>AC,  the  Z  A  MB  will  be  obtuse,  and 
the  Z  AMQ  will  be  acute.  §  152 

Then        AB*  =  BM*  +  AM*  +  2BM X  MD,  §343 

(in  any  obtuse  A  the  square  of  the  side  opposite  the  obtuse  /.  is  equal  to  the 
sum  of  the  squares  of  the  other  two  sides  increased  by  twice  the  product 
of  one  of  those  sides  and  the  projection  of  the  other  on  that  side) ; 


and  AC*  =  MC*  +  AM*-2MCx  MD,  §342 

(in  any  A  the  square  of  the  side  opposite  an  acute  Z.  is  equal  to  the  sum  of 
the  squares  of  the  other  two  sides  diminished  by  twice  the  product  of  one 
of  those  sides  and  the  projection  of  the  other  upon  that  side). 

Add  these  two  equalities,  and  observe  that  BM=  MO. 

Then  AB*  +  AC2  =  2J3M*  +  2AM\ 

Subtract  the  second  equality  from  the  first. 

Then  A^-AO^ZBCx  MD.  Q.E.D. 

NOTE.   This  theorem  enables  us  to  compute  the  lengths  of  the  medians 
if  the  lengths  of  the  three  sides  of  the  triangle  are  known. 


162  PLANE   GEOMETRY.  —  BOOK   III. 


PROPOSITION  XX.    THEOREM. 

345,  If  any  chord  is  drawn  through  a  fixed  point 
within  a  circle,  the  product  of  its  segments  is  con- 
stant in  whatever  direction  the  chord  is  drawn, 


Let  any  two  chords  A3  and  CD  intersect  at  0. 
To  prove  OAxO£=ODx  OC. 

Proof,  Draw  AC  and  ED. 

In  the  A  AOC  and.  BOD, 

Z  0-=  Z.B,  §  263 

(each  being  measured  by  J  arc  AD). 

Z.A  =  /.D,  §  263 

(each  being  measured  by  J  arc  BC}. 

.'.  the  A  are  similar,  §  322 

(two  &  are  similar  when  two  A  of  the  one  are  equal  to  two  A  of  the  other). 

Whence       OA,  the  longest  side  of  the  one, 

:  OD,  the  longest  side  of  the  other, 
:  :  OC,    the  shortest  side  of  the  one, 
:  OB,  the  shortest  side  of  the  other. 

/.  OAxOB=ODx  OC.  §  295 

Q.E.  D. 

346,   SCHOLIUM.     This  proportion  may  be  written 

OA  =  gc      OA     i 

OD     OS  °r  OD~  OB' 
OC 

that  is,  the  ratio  of  two  corresponding  segments  is  equal  to 
the  reciprocal  of  the  ratio  of  the  other  two  corresponding 
segments.  In  this  case  the  segments  are  said  to  be  reciprocally 
proportional. 


NUMERICAL    PROPERTIES   OF    FIGURES.  163 


PROPOSITION  XXL     THEOREM. 

347,  If  from  a  fixed  point  without  a  circle  a  secant 
is  drawn,  the  product  of  the  secant  and  its  external 
segment  is  constant  in  whatever  direction  the  secant 

is  drawn. 

0 


Let  OA  and  OB  be  two  secants  drawn  from  point  0. 

To  prove  OAxOO=OBx  OD. 

Proof.  Draw  BC&nd  AD. 

In  the  A  OAD  and  OBO 

Z.  0  is  common, 
/.A=/-B,  §263 

(each  being  measured  by  J  arc  CD). 

.'.  the  two  A  are  similar,  §  322 

(two  &  are  similar  when  two  A  of  the  one  are  equal  to  two  A  of  the  other). 

Whence         OA,  the  longest  side  of  the  one, 

:  OB,  the  longest  side  of  the  other, 
: :  OD,  the  shortest  side  of  the  one, 
:  OC,  the  shortest  side  of  the  other. 

/.  OA  x  OO=  OB  x  OD.  §  295 

Q.  E.  D. 

REMABK.  The  above  proportion  continues  true  if  the  secant  OB  turns 
about  0  until  B  and  D  approach  each  other  indefinitely.  Therefore,  by 
the  theory  of  limits,  it  is  true  when  B  and  D  coincide  at  H.  Whence, 
OA  x  00=  Off2. 

This  truth  is  demonstrated  directly  in  the  next  theorem. 


164  PLANE   GEOMETRY.  —  BOOK    III. 

PROPOSITION  XXII.     THEOREM. 

348.  If  from  a  point  without  a  circle  a  secant  and 
a  tangent  are  drawn,  the  tangent  is  a  mean  propor- 
tional between  the  whole  secant  and  the  external 
segment. 


Let  OB  be  a  tangent  and  OG  a  secant  drawn  from 
the  point  0  to  the  circle  MBC. 

To  prove  OC\OB  =  OB\  OM. 

Proof,  Draw  BM&nd  BG. 

In  the  A  OEM  and  OBC 

Z  0  is  common. 

Z  OEM  is  measured  by  ^  arc  MB,  §  269 

(being  an  ^.formed  by  a  tangent  and  a  chord). 

Z  (7  is  measured  by  |  arc  EM,  §  263 

(being  an  inscribed  Z). 

.'.  Z  OBM=  Z  G. 

.-.  A  OBC  and  OEM  are  similar,  §  322 

(having  two  A  of  the  one  equal  to  two  A  of  the  other). 

Whence         0(7,    the  longest  side  of  the  one, 

:  OB,  the  longest  side  of  the  other, 
: :  OB,  the  shortest  side  of  the  one, 
:  OM,  the  shortest  side  of  the  other. 

Q.  E.  D. 


NUMERICAL    PROPERTIES   OF    FIGURES.  165 

PROPOSITION  XXIII.     THEOREM. 

349,  The  square  of  the  bisector  of  an  angle  of  a 
triangle  is  equal  to  the  product  of  the  sides  of  this 
angle  diminished  by  the  product  of  the  segments 
determined  by  the  bisector  upon  the  third  side  of  the 
triangle. 


Let  AD  bisect  the  angle  BAG  of  the  triangle  ABC. 

To  prove        AD*  =  AE  X  A  C~  DB  x  DO. 
Proof,      Circumscribe  the  O  ABC  about  the  A  ABC.  §  285 
Produce  AD  to  meet  the  circumference  in  E,  and  draw  EC. 
Then  in  the  A  ABD  and  AEG, 

ZBAD  =  ZCAE,  Hyp. 

Z  B  -  Z  E,  §263 

(each  being  measured  by  J  the  arc  AC). 

.-.  A  ABD  and  AEC  we  similar,  §  322 

(two  &  are  similar  if  two  A  of  the  one  are  equal  respectively  to  two  d  of 
the  other). 

Whence       AB,  the  longest  side  of  the  one, 

:  AE,  the  longest  side  of  the  other, 
:  :  AD,  the  shortest  side  of  the  one, 
:  A  C,  the  shortest  side  of  the  other. 

.-.  AB  x  AC=  ADxAE  §  295 

=  AD(AD+DE) 
=  AD*+ADxDE. 

But  AD  X  DE=  DB  x  DC,  §  345 

(the  product  of  the  segments  of  a  chord  drawn  through  a  fixed  point  in 
aQ  is  constant). 


Whence        AD*  =  ABxAC-DBx  DC.  Q.  E.  D. 

NOTE.  This  theorem  enables  us  to  compute  the  lengths  of  the  bisectors 
of  the  angles  of  a  triangle  if  the  lengths  of  the  sides  are  known. 


166  PLANE   GEOMETRY.  —  BOOK    III. 


PROPOSITION  XXIV.     THEOREM. 

350.  In  any  triangle  the  product  of  two  sides  is 
equal  to  the  product  of  the  diameter  of  the  circum- 
scribed circle  by  the  altitude  upon  the  third  side. 


Let  ABC  be  a  triangle,  AD  the  altitude,  and  ABO 
the  circle  circumscribed  about  the  triangle  ABC. 

Draw  the  diameter  AE,  and  draw  EG. 

To  prove  AB  X  AC=AExAZ>. 

Proof,    In  the  A  ABD  and  AEG, 

Z  EDA  is  a  rt.  Z,  Cons. 

Z  EGA  is  a  rt.  Z,  §  264 

(being  inscribed  in  a  semicircle), 

and  Z  B  -  Z  E.  §  263 

.-.  A  ABD  and  AEG  are  similar,  §  323 

(two  rt.  &  having  an  acute  Z.  of  the  one  equal  to  an  acute,  Z  of  the  other 
are  similar). 

Whence       AB,  the  longest  side  of  the  one, 

:  AE,  the  longest  side  of  the  other, 
:  :  AD,  the  shortest  side  of  the  one, 
:  AG,  the  shortest  side  of  the  other. 

.:.  AB  X  AC=  AE  X  AD.  §  295 

a.  K.  a 

NOTE.  This  theorem  enables  us  to  compute  the  length  of  the  radius  of 
a  circle  circumscribed  about  a  triangle,  if  the  lengths  of  the  three  sides 
of  the  triangle  are  known. 


PROBLEMS    OF   CONSTRUCTION.  167 

PROBLEMS  OF  CONSTRUCTION. 

PROPOSITION  XXV.     PROBLEM. 

351,  To  divide  a  given  straight  line  into  parts  pro- 
portional to  any  number  of  given  lines. 

H        K     7? 


Let  AB,  m,  n,  and  p,  be  given  straight  lines. 
To  divide  AB  into  parts  proportional  to  m,  n,  and  p. 

Construction,     Draw  AX,  making  an  acute  Z  with  AB. 
On  AX  take  AC=m,  CE=n,  EX=p. 

Draw  BX. 
From  ^and  <7draw  EKand  OH  II  to  BX. 

.ZTand  J7are  the  division  points  required. 

Pr    .  (AK\     AH     HK     KB  g 

Pr°°f>  (AEr-AQ—CE—EX' 

(a  line,  drawn  through  two  sides  of  a  A  II  to  the  third  side  divides  those 
sides  proportionally). 

.'.  AH  :  HK  :  KB  =  AC  :  CE  :  EX. 
Substitute  m,  n,  and  p  for  their  equals  AC,  CE,  and  EX. 
Then          AH  :  HK  :  KB  -  m  :  n  :p. 

Q.  E.  F. 


168  PLANE   GEOMETRY.  —  BOOK    III. 


PROPOSITION  XXVI.     PROBLEM. 

352,  To  find  a  fourth  proportional  to  three  given 
straight  lines. 

m          B     n      C  „ 


Let  the  three  given  lines  be  m,  n,  and  p. 

To  find  a  fourth  proportional  to  m,  n,  and  p. 

Draw  Ax  and  Ay  containing  any  acute  angle. 
Construction,  On  Ax  take  AB  equal  to  m,  BC=n. 
On  Ay  take  AD=p. 

Draw  BD. 

From  C  draw  CF  \\  to  BD,  to  meet  Ay  at  F. 
DF  is  the  fourth  proportional  required. 

Proof,  A  B  :  BO  =  AD  :  DF,  §  309 

(a  line  drawn  through  two  sides  of  a  A  II  to  the  third  side  divides  those 
sides  proportionally}. 

Substitute  m,  ?i,  andp  for  their  equals  AB,  BC,  and  AD. 
Then  m  :  n  =p  :  DF. 

Q.  E.  F. 


PROBLEMS   OF   CONSTRUCTION.  169 


PROPOSITION  XXVH.     PROBLEM. 

353,  To  -find    a    third   proportional  to   two  given 
straight  lines. 

A 


Let  m  and  n  be  the  two  given  straight  lines. 

To  find  a  third  proportional  to  m  and  n. 
Construction,     Construct  any  acute  angle  A, 

and  take  AB=m,  AC—n. 
Produce  AB  to  D,  making  ED  =-- AC. 

Join  BO. 

Through  D  draw  DE  II  to  BOio  meet  AC  produced  at  E. 
CE  is  the  third  proportional  to  AB  and  AC. 

Proof,  AB  :  BD  =  A  C :  CE.  §  309 

(a  line  drawn  through  two  sides  of  a  A  II  to  the  third  side  divides  those 
sides  proportionally). 

Substitute,  in  the  above  proportion,  A  C  for  its  equal  BD. 

Then   AB  :  AC=AC:  CE. 

That  is,  m  :  n  —  n  :  CE. 


Q.E.F. 


Ex.  217.  Construct  x,  if  (1)  x  =  — ,  (2)  x  =  -• 

C  C 

Special  Cases  :  (1)  a  =  2,  b  =  3,   c  =  4;    (2)  a  =  3,   6  =  7,   c=ll;(3) 
q,  =  2,  c  =>  3  ;   (4)  a  =  3,  c  =  5 ;   (5)  a  =  2c, 


170  PLANE   GEOMETRY.  —  BOOK    III. 

PROPOSITION  XXVIII.     PROBLEM. 

354,  To  find  a  mean  proportional  between  two  given 
straight  lines. 


m 


Ai£.  ________  .....  ____  L 

m         c 


Let  the  two  given  lines  be  m  and  n. 

To  find  a  mean  proportional  between  m  and  n. 
Construction,        On  the  straight  line  AE 

take  AC~  m,  and  CB  —  n. 

On  AB  as  a  diameter  describe  a  semi-circumference. 
At  C  erect  the  _L  CHis  meet  the  circumference  at  3. 

CHis  a  mean  proportional  between  ra  and  n. 
Proof,  /.  AC  :  CH  =  CH  :  CB,  §  337 

(the  J_  let  fall  from  a  point  in  a  circumference  to  the  diameter  of  a  circle 
is  a  mean  proportional  between  the  segments  of  the  diameter). 

Substitute  for  .4  (7  and  CB  their  equals  in  and  n. 
Then  m  :  CH  =  CH  ;  n. 

Q.  E.  F. 

355,  A  straight  line  is  said  to  be  divided  in  extreme  and 
'mean  ratio,  when  the  whole  line  is  to  the  greater  segment  as 
the  greater  segment  is  to  the  less. 


Ex.  21 8.   Construct  x  if  x  =  Vab. 

Special  Cases  :  (1)  a  =  2,  b  =  3  ;  (2)  a  -  1,  b  =  5  ;  (3)  a  =  3,  6  =  7. 


PROBLEMS   OF   CONSTRUCTION.  171 


v  PROPOSITION  XXIX.     PROBLEM. 

356,  To  divide  a  given  line  in  extreme  and  mean 
ratio. 


C---- 


A  C  B 

Let  AB  be  the  given,  line. 

To  divide  AB  in  extreme  and  mean  ratio. 
Construction,     At  B  erect  a  JL  BE  equal  to  one-half  of  AB. 
From  E  as  a  centre,  with  a  radius  equal  to  EB,  describe  a  O. 
Draw  AE,  meeting  the  circumference  in  .Fand  G. 

On  A3  take  AC  =AF. 
On  BA  produced  take  AC'  ==  AG. 

Then  AB  is  divided  internally  at  G  and  externally  at  C' 
in  extreme  and  mean  ratio. 

Proof,  AG  :  AB  -  AB  :  AF,  §  348 

(if  from  a  point  without  a  O  a  secant  and  a  tangent  are  drawn,  the  tan- 
gent is  a  mean  proportional  between  the  whole  secant  and  the  external 
segment). 

Then  by  §  301  and  §  300, 

AG-AB:AB=AB-AF:  AF,  (1) 

AG  +  AB-.AG^AB  +  AF:  AB.  (2) 

By  construction    FG  =  2  EB  ==  AB. 


Hence  (1)  becomes 

AC:  AB  =  BC  :  AC- 
or,  by  inversion,      AB  :   AC=  AC  :  EG.  §  299 

Again,  since      C'A  =  AG  =  AB  +  AF, 
(2)  becomes  C'JB  :  C'A  =  C'A  :  AB. 

Q.E.F. 


172 


PLANE   GEOMETRY.  —  BOOK    III. 


PROPOSITION  XXX.     PROBLEM. 

357.   Upon  a  given  line  homologous  to  a  given  side 
of  a  given  polygon,  to  construct  a  polygon  similar  to 
the  given  polygon. 
E 


Let  A'E1  be  the  given  line  homologous  to  AE  of  the 
given  polygon  ABODE. 

To  construct  on  A'JE1  a  polygon  similar  to  the  given  polygon. 
Construction,    From  E  draw  the  diagonals  EB  and  EC. 

From  E1  draw  E'B',  E'C',  and  E'D', 
making  A  A'E'B',  B'E'C',  and  C'E'D'  equal  respectively  to 

A  AEB,  EEC,  and  CED. 
From  A'  draw  A'B',  making  Z  E'A'B'  =  Z  EAB, 

and  meeting  E'B1  at  B'. 
From  B1  draw  B'C',  making  Z  E'B'C'  =  Z  EBC, 

and  meeting  E'C'  at  C1. 
From  C1  draw  C'D',  making  Z  E'C'D'  =  Z  ^OZ>, 

and  meeting  E'D'  at  D'. 
Then  A'B'C'D'E'  is  the  required  polygon. 

Proof,    The  corresponding  A  ABE  and  A'B'E',  EEC  and 
E'B'C1,  ECD  and  £"<?'£>'  are  similar,  §  322 

(two  &  are  similar  if  they  have  two  A  of  the  one  equal  respectively  to  two 
A  of  the  other). 

Then  the  two  polygons  are  similar,  §  331 

(two  polygons, composed  of  the  same  number  of  A  similar  to  each  other  and 
similarly  placed,  are  similar). 

Q.E.F, 


PKOBLEMS   OF   COMPUTATION. 


173 


PROBLEMS  OF  COMPUTATION. 

219.   To  compute  the  altitudes  of  a  triangle  ia  terms  of  its  sides. 
O 


('   A    D 

Fm.  1. 

At  least  one  of  the  angles  A  or  B  is  acute.    Suppose  it  is  the  angle  B. 


In  the  A  CDS, 
In  the  A  ABC, 


h*  =  a2  -  BD\ 

^2  =  a2  +  c2  -  2cx  BD. 


§  338 
§342 


Whence, 
Hence 


4  c2  4  c2 

=  (2ac  +  a2  +  c2  -  62)(2ac  -  a2  -  c2  + 
4c2 


Let 
Then 


Hence 


4c2 

_  (a  +  6  +  c)  (a  +  c  —  b)  (b  +  a  —  c)  (6  —  a  +  c) 

4c2 

a  +  b  +  c  =  2s. 
a  +  c  -  6  =  2(s-  b), 


—  a  +  c  =  2(s  —  a). 
-alx  2(,-6)x2(,-oX 
4c2 


By  simplifying,  and  extracting  the  square  root, 

2      

c 

220.   To  compute  the  medians  of  a  triangle  in  terms  of  its  sides. 

By  g  344,  a2  +  &2  =  2m2  +  2/'-Y.  (Fig.  2) 

Whence  4  m2  =  2  (a2  +  62)  -  c2. 


174 


PLANE    GEOMETRY. BOOK    III. 


221.   To  compute  the  bisectors  of  a  triangle  in  terms  of  the  sides. 
By  \  349,        V  =  ab  -  AD  x  BD. 

2010     AD     BD     AD  +  BD        c 

By  I  313,    — —  = = — = -• 

b          a  a  +  b          a  +  b 


Whence 


.-.AD  =  -^-,    and    BD--2*-. 

a  +  b  a  +  b 

abc2 

~~  (a  +  bf 


"Whence 


^  ab  {(a  +  b}2  -  c2} 

(a  +  by 
=  ab  (a  +  b  -f  c)  (a  +  b  —  c) 

a5x2sx2(s-c) 

(a  +  by 

2 


£ 


a  +  b 


222.   To  compute  the  radius  of  the  circle  circumscribed  about  a  tri- 
angle in  terms  of  the  sides  of  the  triangle. 

By  {  350,     AB  X  AC=  AEx  AD, 
or  lc  =  2Rx  AD. 

But 


AD 

Whence        R  - 


=  -  Vs  (s  —  a)  (s  —  b)  (s  —  c). 
a 

abc 


E 


223.  If  the  sides  of  a  triangle  are  3,  4,  and  5,  is  the  angle  opposite  5 
right,  acute,  or  obtuse  ? 

224.  If  the  sides  of  a  triangle  are  7,  9,  and  12,  is  the  angle  opposite 
12  right,  acute,  or  obtuse  ? 

225.  If  the  sides  of  a  triangle  are  7,  9,  and  11,  is  the  angle  opposite 
11  right,  acute,  or  obtuse  ? 

226.  The  legs  of  a  right  triangle  are  8  inches  and  12  inches ;  find  the 
lengths  of  the  projections  of  these  legs  upon  the  hypotenuse,  and  the  dis- 
tance of  the  vertex  of  the  right  angle  from  the  hypotenuse. 

— -  227.  If  the  sides  of  a  triangle  are  6  inches,  9  inches,  and  12  inches, 
find  the  lengths  (1)  of  the  altitudes  ;  (2)  of  the  medians  ;  (3)  of  the  bisec- 
tors ;  (4)  of  the  radius  of  the  circumscribed  circle. 


EXERCISES.  175 


THEOREMS. 

228.  Any  two  altitudes  of  a  triangle  are  inversely  proportional  to 
the  corresponding  bases. 

229.  Two  circles  touch  at  P.     Through  P  three  lines  are  drawn,  meet- 
ing one  circle  in  A,  B,  C,  and  the  other  in  A',  B' ,  (7,  respectively.   Prove 
that  the  triangles  ABC,  A/£/C/  are  similar. 

'230.  Two  chords  AB,  CD  intersect  at  M,  and  A  is  the  middle  point  of 
the  arc  CD.  Prove  that  the  product  AB  x  AM  remains  the  same  if  the 
chord  AB  is  made  to  turn  about  the  fixed  point  A. 

HINT.  Draw  the  diameter  AE,  join  BE,'qp.&  compare  the  triangles 
thus  formed. 

231.  The  sum  of  the  squares  of  the  segments  of  two  perpendicular 
chords  is  equal  to  the  square  of  the  diameter  of  the  circle. 

HINT.  If  AB,  CD  are  the  chords,  draw  the  diameter  BE,  join  AC, 
ED,  BD,  and  prove  that  AC  =  ED.  Apply  §  338. 

232.  In  a  parallelogram  ABCD,  a  line  DE  is  drawn,  meeting  the 
diagonal  AC  in  F,  the  side  BC  in  G,  and  the  side  AB  produced  in  E. 
Prove  that  DF2  =  FGx  FE. 

233.  The  tangents  to  two  intersecting  circles  drawn  from  any  point 
in  their  common  chord  produced,  are  equal.     (§  348.) 

234.  The  common  chord  of  two  intersecting  circles,  if  produced,  will 
bisect  their  common  tangents.     (§  348.) 

7  235.   If  two  circles  touch  each  other,  their  common  tangent  is  a  mean 
proportional  between  their  diameters. 

HINT.  Let  AB  be  the  common  tangent.  Draw  the  diameters  AC,  BD. 
Join  the  point  of  contact  P  to  A,  B,  0,  and  D.  Show  that  APD  and  BPC 
are  straight  lines  JL  to  each  other,  and  compare  A  ABC,  ABD. 

236.  If  three  circles  intersect  one  another,  the  common  chords  all  pass 
-$hrough  the  same  point. 

HINT.  Let  two  of  the  chords  AB  and  CD 
meet  at  0.  Join  the  point  of  intersection  E 
to  0,  and  suppose  that  EO  produced  meets 
the  same  two  circles  at  two  different  points  P 
and  Q.  Then  prove  that  OP=  0Q;  hence, 
that  the  points  P  and  Q  coincide, 


176 


PLANE   GEOMETRY.  —  BOOK    Til. 


237.  If  two  circles  are  tangent  internally,  all  chords  of  the  greater 
circle  drawn  from  the  point  of  contact  are  divided  proportionally  by  the 
circumference  of  the  smaller  circle. 

HINT.   Draw  any  two  of  the  chords,  join  the  points  where  they  meet 
the  circumferences,  and  prove  that  the  A  thus  formed  are  similar. 

238.  In  an  inscribed  quadrilateral,  the  product  of  the  diagonals  is 
equal  to  the  sum  of  the  products  of  the  opposite  sides. 

HINT.  Draw  DE,  making  Z.  CDE  =  Z.ADB.  The 
&  ABD  and  CDE  are  similar.  Also  the  &  BCD  and 
ADE  are  similar. 

239.  The  sum  of  the  squares  of  the  four  sides  of 
any  quadrilateral  is  equal  to  the  sum  of  the  squares 
of  the  diagonals,  increased  by  four^imes  the  square 
of  the  line  joining  the  middle  points  of  the  diagonals. 

HINT.  Join  the  middle  points  F,  E,  of  the  diag- 
onals. Draw  EB  and  ED.  Apply  §  344  to  the 
A  ABC  and  ADC,  add  the  results,  and  eliminate 
BE2  +  DE'2  by  applying  g  343  to  the  A  BDE. 

\Jr\ 

240.  The  square  of  the  bisector  or  an  exterior  angle  of  a  triangle  is 

equal  to  the  product  of  the  external  segments  deter- 
mined by  the  bisector  upon  one  of  the  sides,  dimin- 
ished by  the  product  of  the  other  two  sides. 

HINT.   Let  CD  bisect  the  exterior  Z  BCH  of 
the  A  ABC.    Circumscribe  a  O  about  the  A,  pro- 
duce DC  to  meet  the  circumference  in  F,  and  draw  BF.    Prove  &ACD, 
BCF  similar.    Apply  g  347. 

<-  —  j  241.   If  a  point  0  is  joined  to  the  vertices  of  a  triangle  ABC,  and 

/through  any  point  A/  in  OA  a  line  parallel  to  AB  is  drawn,  meeting  OB 

at  B',  and  then  through  B'  a  line  parallel  to  EC,  meeting  00  at  (7, 

and  C/  is  joined  to  A',  the  triangle  A'B'C'  will  be  similar  to  the  tri- 

angle ABC. 

-"'7     242.   If  the  line  of  centres  of  two  circles  meets  the  circumferences  at 
/  the  points  A,  B,  (7,  D,  and  meets  the  common  exterior  tangent  at  P,  then 


D 


71 


243.  The  line  of  centres  of  two  circles  meets  the  common  exterior 
tangent  at  P,  and  a  secant  is  drawn  from  P,  cutting  the  circles  at  the 
consecutive  points  E,  F,  G,  H.  Prove  that  PExPH--=  PFx  PQ. 


EXERCISES.  177 


NUMERICAL  EXERCISES. 

TP  244.  A  line  is  drawn  parallel  to  a  side  AB  of  a  triangle  ABC,  and 
Cutting  AC  in  D,  BC  in  E,  If  AD  :  DC=  2  :  3,  and  AB=  20  inches, 
find  D#  • 


245.   The  sides  of  a  triangle  are  9, 12, 15.    Find  the  segments  made  by 
bisecting  the  angles.    (§  313.) 

^2  246.   A  tree  casts  a  shadow  90  feet  long,  when  a  vertical  rod  6  feet 
High  casts  a  shadow  4  feet  long.     How  high  is  the  tree  ? 

247.   The  bases  of  a  trapezoid  are  represented  by  a,  6,  and  the  altitude 
ft.     Find  the  altitudes  of  the  two  triangles  formed  by  producing  the 
legs  till  they  meet. 

VV248.   The  sides  of  a  triangle  are  6,  7,  8.    In  a  similar  triangle  the  side 
homologous  to  8  is  equal  to  40.     Find  the  other  two  sides. 

249.  The  perimeters  of  two  similar  polygons  are  200  feet  and  300  feet. 
If  a  side  of  the  first  polygon  is  24  feet,  find  the  homologous  side  of  the 
second  polygon. 

250.  How  long  must  a  ladder  be  to  reach  a  window  24  feet  high,  if 
wie  lower  end  of  the  ladder  is  10  feet  from  the  side  of  the  house  ? 

251.  If  the  side  of  an  equilateral  triangle  =  a,  find  the  altitude. 

252.  If  the  altitude  of  an  equilateral  triangle  =  A,  find  the  side. 

;  253.  Find  the  lengths  of  the  longest  and  the  shortest  chord  that  can 
bo  drawn  through  a  point  6  inches  from  the  centre  of  a  circle  whose 
radius  is  equal  to  10  inches. 

254.  The  distance  from  the  centre  of  a  circle  to  a  chord  10  inches  long 
is  12  inches.   Find  the  distance  from  the  centre  to  a  chord  24  inches  long. 

255.  The  radius  of  a  circle  is  5  inches.    Through  a  point  3  inches  from 
tho  centre  a  diameter  is  drawn,  and  also  a  chord  perpendicular  to  the 
diameter.     Find  the  length  of  this  chord,  and  the  distance  from  one  end 
of  the  chord  to  the  ends  of  the  diameter. 

256.  The  radius  of  a  circle  is  6  inches.     Through  a  point  10  inches 
from  the  centre  tangents  are  drawn.     Find  the  lengths  of  the  tangents, 
and  also  of  the  chord  joining  the  points  of  contact. 

257.  If  a  chord  8  inches  long  is  3  inches  distant  from  the  Centre  of 
the  circle,  find  the  radius  and  the  distances  from  the  end  of  the  chord  to 
the  ends  of  the  diameter  which  bisects  the  chord. 


178  PLANE   GEOMETRY.  —  BOOK   III. 

258.  The  radius  of  a  circle  is  13  inches.  Through  a  point  5  inches 
f^6m  the  centre  any  chord  is  drawn.  What  is  the  product  of  the  two  seg- 
ments of  the  chord  ?  What  is  the  length  of  the  shortest  chord  that  can 
be  drawn  through  the  point  ? 

*J     259.   From  the  end  of  a  tangent  20  inches  long  a  secant  is  drawn 
f  through  the  centre  of  the  circle.     If  the  exterior  segment  of  this  secant 
is  8  inches,  find  the  radius  of  the  circle. 

2GO.  The  radius  of  a  circle  is  9  inches  ;  the  length  of  a  tangent  is  12 
inches.  Find  the  length  of  a  secant  drawn  from  the  extremity  of  the 
tangent  to  the  centre  of  the  circle. 

261.  The  radii  of  two  circles  are  8  inches  and  3  inches,  and  the  dis- 
tance between  their  centres  is  15  inches.  Find  the  lengths  of  their  com- 
mon tangents. 

-262.  Find  the  segments  of  a  line  10  inches  long  divided  in  ex4reme 
and  mean  ratio. 

'263.  The  sides  of  a  triangle  are  4,  5,  6.  Is  the  largest  angle  acute, 
right,  or  obtuse  ? 

PROBLEMS. 

264.  To  divide  one  side  of  a  given  triangle  into  segments  proportional 
to  the  adjacent  sides.     (§  313.) 

265.  To  produce  a  line  AB  to  a  point  C so  that  AB  :  AC=  3  :  5. 

266.  To  find  in  one  side  of  a  given  triangle  a  point  whose  distances 
from  the  other  sides  shall  be  to  each  other  in  a  given  ratio. 

267.  Given  an  obtuse  triangle  ;  to  draw  a  line  from  the  vertex  of  the 
obtuse  angle  to  the  opposite  side  which  shall  be  a  mean  proportional 
between  the  segments  of  that  side. 

268.  Through  a  given  point  P  within  a  given  circle  to  draw  a  chord 


269.  To  draw  through  a  given  point  P  in  the  arc  subtended  by  a  chord 
AB  a  chord  which  shall  be  bisected  by  AB. 

270.  To  draw  through  a  point  P,  exterior  to  a  given  circle,  a  secant 
PAB  so  that  PA :  AB  =  4 :  3. 

271.  To  draw  through  a  point  P,  exterior  to  a  given  circle,  a  secant 
PAB  so  that  Iff  =  PA  X  PB. 

272.  To  find  a  point  P  in  the  arc  subtended  by  a  given  chord  AB  so 


EXERCISES.  179 

273.  To  draw  through  one  of  the  points  of  intersection  of  two  circles 
a  secant  so  that  the  two  chords  that  are  formed  shall  be  to  each  other 
in  the  ratio  of  3  :  5. 

274.  To  divide  a  line  into  three  parts  proportional  to  2,  f  ,  J. 

275.  Having  given  the  greater  segment  of  a  line  divided  in  extreme 
and  mean  ratio,  to  construct  the  line. 

276.  To  construct  a  circle  which  shall  pass  through  two  given  points 
and  touch  a  given  straight  line. 

277.  To  construct  a  circle  which  shall  pass  through  a  given  point  and 
touch  two  given  straight  lines. 

278.  To  inscribe  a  square  in  a  semicircle. 

279.  To  inscribe  a  square  in  a  given  triangle. 

HINT.  Suppose  the  problem  solved,  and  DEFQ-  the  inscribed  square. 
Draw  CM  II  to  AB,  and  let  AF  produced  meet 
CM  in  M.  Draw  Off  and  MN  ±  to  AB,  and 
produce  AB  to  meet  MN  at  N.  The  A  ACM, 
AGF  are  similar;  also  the  &  AMN,  AFE 
are  similar.  By  these  triangles  show  that 
the  figure  CMNH  is  a  square.  By  construct- 
ing this  square,  the  point  F  can  be  found. 

280.  To  inscribe  in  a  given  triangle  a  rectangle  similar  to  a  given 
rectangle. 

281.  To  inscribe  in  a  circle  a  triangle  similar  to  a  given  triangle. 

282.  To  inscribe  in  a  given  semicircle  a  rectangle  similar  to  a  given 
rectangle. 

283.  To  circumscribe  about  a  circle  a  triangle  similar  to  a  given 
triangle. 


284.  To  construct  the  expression,  x  =    -       ;  that  is,  —  X  -• 

de  d       e 

285.  To  construct  two  straight  lines,  having  given  their  sum  and 
their  ratio. 

286.  To  construct  two  straight  lines,  having  given  their  difference 
and  their  ratio.  w;  "  . 

287.  Having  given  two  circles,  with  c.4re$9  0  and  0/,  and  a  point  A 
in  their  plane,  to  draw  through  the  point  A  a  straight  line,  meeting  the 
circumferences  at  B  and  Ct  so  that  AB  :  AC=  1  :  2. 

HINT.   Suppose  the  problem  solved,  join  OA  and  produce  it  to  Dt 
making  OA  :  AD  «  1  :  2.     Join  DC;  &  OA  B,  ADO  are  similar. 


BOOK   IV. 


AREAS    OF    POLYGONS. 

358,  The  area  of  a  surface  is  the  numerical  measure  of  the 
surface  referred  to  the  unit  of  surface. 

The  unit  of  surface  is  a  square  whose  side  is  a  unit  of  length  ; 
as  the  square  inch,  the  square  foot,  etc. 

359,  Equivalent  figures  are  figures  having  equal  areas. 

PEOPOSITION  I.    THEOREM. 

360,  The  areas  of  two  rectangles  having  equal  alti- 
tudes are  to  each  other  as  their  bases. 


\ 


D 


0 


Let  the  two  rectangles  he  AC  and  AF,  having  the 
same  altitude  AD. 

m  rect.  AC     AB 

To  prove  •         =  —-• 

Proof,    CASE  I.    When  AB  and  AE  are  commensurable. 

Suppose  AB  and  AE  have  a  common  measure,  as  AO, 

which  is  contained  in  AB  seven  times  and  in  AE  four  times. 

AB  _7 

AE      4 

Apply  this  measure  to  AB  and  AE,  and  at  the  several 
points  of  division  erect  Js. 

The    rect.  AC  will  be  divided  into  seven  rectangles, 
and  the  rect.  A F  will  be  divided  into  four  rectangles. 


Then 


(1) 


AREAS  OF   POLYGONS.  181 

These  rectangles  are  all  equal.  §  186 


Hence  T^=T  (2) 

lect.AF     4 


R-W-dfl)    gfirft  A,., 

CASE  II.    TFAeft  ^.^  arcc?  J.^57  are  incommensurable. 


1(7        D 


H 


F 


K 


Divide  AB  into  any  number  of  equal  parts,  and  apply  one 
of  them  to  AE  as  often  as  it  will  be  contained  in  AE. 

Since  AB  and  AE  are  incommensurable,  a  certain  number 
of  these  parts  will  extend  from  A  to  a  point  K,  leaving  a 
remainder  KE  less  than  one  of  the  parts. 
Draw  KH II  to  EF. 

Since  AB  and  AK  &IQ  commensurable, 

recD.  j-j-jLi      -/TL-ti.  /**       f 

777  =  "T5"  Case! 

rect.  AC      AB 

These  ratios  continue  equal,  as  the  unit  of  measure  is  indefi- 
nitely diminished,  and  approach  indefinitely  the  limiting  ratios 

Teci.AF      -,  AE 

—  and  -  -  respectively. 

rect.  AC          AB 

-  rect.  AF=  AE  ,  280 

"rect.  AC      AB' 

(if  two  variables  are  constantly  equal,  and  each  approaches  a  limit,  the 
limits  are  equal).  Q  E>  D 

361.  COR.  The  areas  of  two  rectangles  having  equal  bases  are 
to  each  other  as  their  altitudes.  For  AB  and  AE  may  be  con- 
sidered as  the  altitudes,  AD  and  AD  as  the  bases. 

NOTE.  In  propositions  relating  to  areas,  the  words  "  rectangle," 
"triangle,"  etc.,  are  often  used  for  "area  of  rectangle,"  "area  of  tri- 
angle," etc. 


182 


PLANE   GEOMETRY.  —  BOOK    IV. 


PROPOSITION  II.     THEOREM. 

362,  The  areas  of  two  rectangles  are  to  each  other 
as  the  products  of  their  bases  by  their  altitudes. 


b  b'  b 

Let  R  and  E'  be  two  rectangles,  having  for  their 
bases  b  and  b',  and  for  their  altitudes  a  and  a1. 
E  =  aX  b 
R' 


To  prove 


a'xb' 

Proof,    Construct  the  rectangle  S,  with  its  base  the  same  as 
that  of  JR,  and  its  altitude  the  same  as  that  of  E]. 

E     a 


Then  4H-,'  §361 

S     a1 

(rectangles  having  equal  bases  are  to  each  other  as  their  altitudes) ; 

and  -!.  =  !:.  §360 


(rectangles  having  equal  altitudes  are  to  each  other  as  their  bases). 
By  multiplying  these  two  equalities, 


lt_       a  X  b 

E1      a1  X  b'' 


Q.  E.  o. 


Ex.  288.  Find  the  ratio  of  a  rectangular  lawn  72  yards  by  49  yards 
to  a  grass  turf  18  inches  by  14  inches. 

Ex.  289.  Find  the  ratio  of  a  rectangular  courtyard  18 J  yards  by  15J 
yards  to  a  flagstone  31  inches  by  18  inches. 

Ex.  290.  A  square  and  a  rectangle  have  the  same  perimeter,  100  yards. 
The  length  of  the  rectangle  is  4  times  its  breadth.  Compare  their  areas. 

Ex.  291.  On  a  certain  map  the  linear  scale  is  1  inch  to  Smiles.  How 
many  acres  are  represented  on  this  map  by  a  square  the  perimeter  of 
which  is  1  inch  ? 


AREAS   OF   POLYGONS. 


183 


PROPOSITION  III.     THEOREM. 

L/ 

363,  The  area  of  a  rectangle  is  equal  to  the  product 

of  its  base  and  altitude. 


B 


Let  E  be  the  rectangle,  b  the  base,  and  a  the  alti- 
tude; and  let  U  be  a  square  whose  side  is  equal  to 
the  linear  unit. 

To  prove  the  area  of  E  =  a  X  b. 


It      a  X  b  ,  , 

~  =  - . -  =  aX  £, 

Z7     1x1 


§362 


(two  rectangles  are  to  each  other  as  the  product  of  their  bases  and  altitudes). 


But 


—  =  the  area  of  JR. 


358 


.*.  the  area  of  R  =  a  X  b.  Q>  E<  Dm 

364,  SCHOLIUM.  When  the  base  and  altitude  each  contain 
the  linear  unit  an  integral  number  of  times,  this  proposition  is 
rendered  evident  by  dividing  the  figure  into  squares,  each 


equal  to  the  unit  of  measure.  Thus,  if  the  base  contain  seven 
linear  units,  and  the  altitude  four,  the  figure  may  be  divided 
into  twenty-eight  squares,  each  equal  to  the  unit  of  measure ; 
and  the  area  of  the  figure  equals  7x4  units  of  surface. 


184  PLANE   GEOMETRY.  —  BOOK   IV. 


PROPOSITION  IV.     THEOREM. 

365,  The  area  of  a  parallelogram,  is  equal  to  the 
product  of  its  base  and  altitude. 
B    IE C      F 


I)  D  A         b         D 

Let  AEFD  "be  a  parallelogram,  AD  its  base,  and  CD 
its  altitude. 

To  prove  the  area  of  the  O  AEFD  =  AD  X  CD. 
Proof,   From  A  draw  AB  II  to  DO  to  meet  FE  produced. 
Then  the  figure  ABCD  will  be  a  rectangle,  with  the  same 
base  and  altitude  as  the  O  AEFD. 
In  the  rt.  A  ABE  and  DCF 

AB^  CD  and  AE=  DF,  §  179 

(being  opposite  sides  of  a  O). 

.'.AAJSE=ADCFJ  §161 

(two  rt.  A  are  equal  when  the  hypotenuse  and  a  side  of  the  one  are  equal 
respectively  to  the  hypotenuse  and  a  side  of  the  other). 

Take  away  the  A  DCF,  and  we  have  left  the  rect.  ABCD. 
Take  away  the  A  ABE,  and  we  have  left  the  O  AEFD. 

.'.  rect.  ABCD  =c=  O  AEFD.  Ax.  3 

But  the  area  of  the  rect,  ABCD  =  a  X  b,         §  363 

.*.  the  area  of  the  O  AEFD  =  axb.  Ax.  1 

Q.  E.  D. 

366,  COR.  1.  Parallelograms  having  equal  bases  and  equal 
altitudes  are  equivalent. 

367,  COR.  2.  Parallelograms  having  equal  bases  are  to  each 
other  as  their  altitudes ;  parallelograms  having  equal  altitudes 
are  to  each  other  as  their  bases ;  any  two  parallelograms  are 
to  each  other  as  the  products  of  their  bases  by  their  altitudes. 


AREAS   OF   POLYGONS. 


185 


PROPOSITION  V.     THEOREM. 


368,  The  area  of  a  triangle  is  equal  to  one-half 
the  product  of  its  base  by  its  altitude. 


D 


Let  ABC  be   a    triangle,    AS  its   base,    and  DC  its 
altitude. 

To  prove  the  area  of  the  A  ABC=  %  AB  X  DQ. 
Proof,  From  O  draw  CH  II  to  BA. 

From  A  draw  AH  II  to  BC. 

The  figure  ABCHis  a  parallelogram, 
(having  its  opposite  sides  parallel), 

and  AC  is  its  diagonal. 


§168 


§178 

(the  diagonal  of  a  ED  divides  it  into  two  equal  A). 

The  area  of  the  O  ABCH  is  equal  to  the  product  of  its 
base  by  its  altitude.  §  365 

Therefore  the  area  of  one-half  the  O,  that  is,  the  area  of 
the  A  ABC,  is  equal  to  one-half  the  product  of  its  base  by  its 
altitude. 

Hence,     the  area  of  the  A  ABC '=  %AB  X  DC. 

Q.  E.  D. 

369.  COR.  1.    Triangles  having  equal  bases  and  equal  alti- 
tudes are  equivalent. 

370.  COR.  2.    Triangles  having  equal  bases  are  to  each  other 
as  their  altitudes ;  triangles  having  equal  altitudes  are  fo  each 
other  as  their  bases ;  any  two  triangles  are  to  each  other  as  the 
products  of  their  bases  by  their  altitudes. 


186 


PLANE   GEOMETRY. —  BOOK    IV. 


PROPOSITION  VI.    THEOREM. 

371,  The  area  of  a  trapezoid  is  equal  to  one-half 
the  sum  of  the  parallel  sides  multiplied  by  the  alti- 
tude. TT  E  ~h' 


.A  F  b  B 

Let  ABCH  be  a  trapezoid,  and  EF  the  altitude. 

To  prove         area  of  ABCH  '=  %  (HC+  AB)  EF 

Proof.  Draw  the  diagonal  AC. 

Then  the  area  of  the  A  ABC=  %  (AB  X  EF),  §  368 

and  the  area  of  the  A  AHC=  \  (ITCx  EF). 

By  adding,     area  of  ABCH=  \  (AB  +  HO)  EF.       Q.  E.  D. 

372,  COB.  The  area  of  a  trapezoid  is  equal  to  the  product 
of  the  median  by  the  altitude.  For,  by  §  191,  OP  is  equal  to 
and  hence 


the  area  of  ABCH=  OP  X  EF. 

373,  SCHOLIUM.  The  area  of  an  irregular  polygon  may  be 
found  by  dividing  the  poly- 
gon into  triangles,  and  by 
finding  the  area  of  each  of 
these  triangles  separately. 
But  the  method  generally 
employed  in  practice  is  to 
draw  the  longest  diagonal, 


F 


and  to  let  fall  perpendiculars  upon  this  diagonal  from  the 
other  angular  points  of  the  polygon. 

The  polygon  is  thus  divided  into  right  triangles  and  trape- 
zoids;  the  sum  of  the  areas  of  these  figures  will  be  the1  area 
of  the  polygon. 


AKEAS   OF   POLYGONS. 


187 


PROPOSITION  VII.     THEOREM. 

374,  The  areas  of  two  triangles  which  have  an  angle 
of  the  one  equal  to  an  angle  of  the  other  are  to  each 
other  as  the  products  of  the  sides  including  the  equal 

angles. 

A 


Let  the  triangles  ABC  and  ADE  have  the  common 
angle  A. 

A  ABC      ABxAC 


To  prove 

Proof, 

Now 


ADxAE 
Draw  BE. 


and 


A  ABC 
AABE= 

A  ABE 


.AC 
AE 

AB 


A  ADE     AD 

(&  having  the  same  altitude  are  to  each  other  as  their  bases). 
By  multiplying  these  equalities, 

A ABC  = ABxAC 
A  ADE     AD  x  AE 


§370 


Q.  E.  D. 


Ex.  292.  The  areas  of  two  triangles  which  have  an  angle  of  the  one 
supplementary  to  an  angle  of  the  other  are  to  each  other  as  the  products 
of  the  sides  including  the  supplementary  angles. 


188 


PLANE   GEOMETEY.  —  BOOK    IV. 


COMPARISON   OF   POLYGONS. 

PROPOSITION  VIII.    THEOREM. 

375,  The  areas  of  two  similar  triangles  are  to  each 
other  as  the  squares  of  any  two  homologous  sides. 


A  O 


A!-  0 


Let  the  two  triangles  be  ACB  and 

&ACB 


To  prove 


A  A'C'JB' 


Draw  the  perpendiculars  CO  and  C'O1. 

A  ACB        ABxCO        AB 
Inen 


X 


CO 


§370 


A  A'C'JP     A'J3'  X  C'O'     A'£'  "  C'&' 
(two  &  are  to  each  other  as  the  products  of  their  bases  by  their  altitudes). 

AB       CO 


But 


A'£'      C'O1 


§328 


(the  homologous  altitudes  of  similar  A  have  the  same  ratio  as  their  homolo- 
gous bases). 

CO  AB 

Substitute,  in  the  above  equality,  for  •         its  equal  ; 

0    C/  J~L  -D 


then 


A  ACB 


AB  xx  AB 

•  x 


AJ? 


A'£'     A'£' 


Q.  E.  D. 


COMPARISON  OF  POLYGONS. 


189 


PROPOSITION  IX.     THEOREM. 

376,  The  areas  of  two  similar  polygons  are  to  each 
other  as  the  squares  of  any  two  homologous  sides. 
E 


B  C  Bf  C' 

Let  S  and  S'  denote  the  areas  of  the  two  similar 
polygons  ABC  etc.,  and  A'B'C'  etc. 

To  prove  S  :  S'  =  A£?  :  1^. 

Proof.  By  drawing  all  the  diagonals  from  the  homologous 
vertices  E  and  E\  the  two  similar  polygons  are  divided  into 
triangles  similar  and  similarly  placed.  §  332 

A  ABE  ' 


.    ? 

(similar  &  are  to  each  other  as  the  squares  of  any  two  homologous  sides). 
Th  A  ABE    _  A  BCE  „  .   A  CPE 

1S> 


&ABE+BCE+CDE 
'E'  +  £'C'E'  +  C'D'E' 


A  ABE 


A^JB?      ^'§3°3 

(in  a  smes  of  equal  ratios  the  sum  of  the  antecedents  is  to  the  sum  of  the 
consequents  as  any  antecedent  is  to  its  consequent). 

.*.#:#  —  AB  :  A1  B' .  Q  E  a 

377,  COR.  1.   The  areas  of  two  similar  polygons  are  to  each 
other  as  the  squares  of  any  two  homologous  lines. 

378,  COR.  2.    The  homologous  sides  of  two  similar  polygons 
have  the  same  ratio  as  the  square  roots  of  their  areas. 


190 


PLANE   GEOMETBY. —  BOOK    IV. 


PROPOSITION  X.     THEOREM. 

379.  The  square  described  on  the  hypotenuse  of  a 
right  triangle  is  equivalent  to  the  sum  of  the  squares 
on  the  other  two  sides. 

Let  BE,  CH,  AF,  be  squares  on  the  three  sides  of  the 
right  triangle  ABC. 


to 


To  prove  £C2  =c»  A$  +  AC*. 

Proof,    Through  A  draw  AL  I 
CE,  and  draw  AD  and  FO. 

Since  A  BAG,  BAG,  and  CAH 
are  rt.  A,  CAG  and  BAH  are 
straight  lines. 

Since  BD  =  BC,  being  sides  of 
the  same  square,  and  BA  =  BF, 
for  the  same  reason,  and  since 
Z  ABD  =  Z.FBC,  each  being  the 
sum  of  a  rt.  Z.  and  the  Z  ABC, 


Now 


E 


§150 


the  rectangle  BL  is  double  the  A  ABD, 
(having  the  same  base  BD,  and  the  same  altitude,  the  distance  between  the 
\\s  AL  and 


and  the  square^4.Fis  double  the  A  FBC, 

(having  the  same  base  FB,  and  the  same  altitude,  the  distance  between  the 
\\s  FB  and  GO). 

Hence  the  rectangle  BL  is  equivalent  to  the  square  AF. 

In  like  manner,  by  joining  AE  and  BK,  it  may  be  proved 
that  the  rectangle  CL  is  equivalent  to  the  square  CH. 

Therefore  the  square  BE,  which  is  the  sum  of  the  rectangles 
BL  and  CL,  is  equivalent  to  the  sum  of  the  squares  CH  and 


380,  COR.  The  square  on  either  leg  of  a  right  triangle  is 
equivalent  to  the  difference  of  the  squares  on  the  hypotenuse  and 
the  other  leg. 


COMPARISON   OF   POLYGONS. 


191 


Ex.  293.  The  square  constructed  upon  the  sum  of  two  straight  lines 
is  equivalent  to  the  sum  of  the  squares  constructed  upon  these  two  lines, 
increased  by  twice  the  rectangle  of  these  lines. 

Let  AB  and  BC  be  the  two  straight  lines,  and  AC  their  sum.     Con- 
struct the  squares  ACGK  and  ABED  upon  AC  and 
AB  respectively.     Prolong  BE  and  DE  until  they 
meet  KG  and  CG  respectively.     Then  we  have  the 
square  EFGH,  with  sides  each  equal  to  BC.      Hence, 
the  square  ACGK  is  the  sum  of  the  squares  ABED  D  - 
and  EFGH,  and  the  rectangles  DEHK  and  BCFE, 
the  dimensions  of  which  are  equal  to  AB  and  BC. 


B    C 


K 


H    G 


H     K 


Ex.  294.  The  square  constructed  upon  the  difference  of  two  straight 
lines  is  equivalent  to  the  sum  of  the  squares  constructed  upon  these  two 
lines,  diminished  by  twice  the  rectangle  of  these  lines. 

Let  AB  and  AC  be  the  two  straight  lines,  and  BC  their  difference. 
Construct  the  square  ABFG  upon  AB,  the  square 
ACKH upon.  AC,  and  the  square  BEDC  upon  BC  (as 
shown  in  the  figure).     Prolong  ED  until  it  meets  AG 
in  L. 

The  dimensions  of  the  rectangles  LEFG  and  HKDL 
are  AB  and  AC,  and  the  square  BCDE  is  evidently 
the  difference  between  the  whole  figure  and  the  sum 
of  these  rectangles ;  that  is,  the  square  constructed  G  F 

upon  BC  is  equivalent  to  the  sum  of  the  squares  constructed  upon  AB 
and  AC  diminished  by  twice  the  rectangle  of  AB  and  AC. 

Ex.  295.  The  difference  between  the  squares  constructed  upon  two 
straight  lines  is  equivalent  to  the  rectangle  of  the  sum  and  difference  of 
these  lines. 

Let  ABDE  and  BCGF  be  the  squares  constructed  upon   the  two 


E 


straight  lines  AB  and  BC.  The  difference  between 
these  squares  is  the  polygon  ACGFDE,  which  poly- 
gon, by  prolonging  CG  to  jET,  is  seen  to  be  composed  of 
the  rectangles  ACHE  and  GFDH.  Prolong  AE  and 
CHto  Jand  .^respectively,  making  ^J/and  UK  each 
equal  to  BC,  and  draw  IK.  The  rectangles  GFDH 
and  EHKI  are  equal.  The  difference  between  the 
squares  ABDE  and  BCGF  is  then  equivalent  to  the 
rectangle  ACKI,  which  has  for  dimensions  AI 
=  AB-  BC. 


K 


G 


H 


F 


A          C     B 

AB  +  BC,  and  EH 


192 


PLANE   GEOMETRY. — BOOK   IV. 


PROBLEMS  OF  CONSTRUCTION. 

PROPOSITION  XI.     PROBLEM. 

381,  To  construct  a  square  equivalent  to  the  sum 
of  two  given  squares. 


B 

Rr 

A 

*».% 

s 

Let  R  and  E1  be  two  given  squares. 

To  construct  a  square  equivalent  to  JR1  +  R. 
Construction,          Construct  the  rt.  Z  A. 

Take  A  0  equal  to  a  side  of  .#', 

AB  equal  to  a  side  of  R\  and  draw  BC. 

Construct  the  square  S,  having  each  of  its  sides  equal  to  BO. 

8  is  the  square  required. 
Proof,  W  =0=  AC2  +  AB\  §  379 


(the  square  on  the  hypotenuse  of  a  rt.  A  is  equivalent  to  the  sum  of  the 
squares  on  the  two  sides). 


Q.  E.  F. 


Ex.  296.  If  the  perimeter  of  a  rectangle  is  72  feet,  and  the  length  is 
equal  to  twice  the  width,  find  the  area. 

Ex.  297.  How  many  tiles  9  inches  long  and  4  inches  wide  will  be 
required  to  pave  a  path  8  feet  wide  surrounding  a  rectangular  court  120 
feet  long  and  36  feet  wide  ? 

Ex.  298.  The  bases  of  a  trapezoid  are  16  feet  and  10  feet;  each  leg 
is  equal  to  5  feet.  Find  the  area  of  the  trapezoid. 


PEOBLEMS    OF   CONSTRUCTION. 


193 


PROPOSITION  XII.     PROBLEM. 

382,  To  construct  a  square  equivalent  to  the  differ- 
ence of  two  given  squares. 


s   ! 


Let  R  be  the  smaller  square  and  R'  the  larger. 

To  construct  a  square  equivalent  to  K  —  It. 
Construction,          Construct  the  rt.  Z  A. 

Take  AB  equal  to  a  side  of  JR. 
From  B  as  a  centre,  with  a  radius  equal  to  a  side  of  R\ 

describe  an  arc  cutting  the  line  ^4JTat  O. 

Construct  the  square  8,  having  each  of  its  sides  equal  to  A  C. 

S  is  the  square  required. 

Proof,  AC2  =0=  B(?  -  AB' \  §  380 

(the  square  on  either  leg  of  a  rt.  A  is  equivalent  to  the  difference  of  the 
squares  on  the  hypotenuse  and  the  other  leg). 

*•£•*' JP-ilfc 

Q.  E.  F. 


Ex.  299.  Construct  a  square  equivalent  to  the  sum  of  two  squares 
whose  sides  are  3  inches  and  4  inches. 

Ex.  300.  Construct  a  square  equivalent  to  the  difference  of  two 
squapis  whose  sides  are  2£  inches  and  2  inches. 

Ex.  301.  Find  the  side  of  a  square  equivalent  to  the  sum  of  two 
squares  whose  sides  are  24  feet  and  32  feet. 

Ex.  302.  Find  the  side  of  a  square  equivalent  to  the  difference  of  two 
squares  whose  sides  are  24  feet  and  40  feet. 

Ex.  303.  A  rhombus  contains  100  square  feet,  and  the  length  of  one 
diagonal  is  10  feet.  Find  the  length  of  the  other  diagonal. 


194  PLANE   GEOMETEY.  —  BOOK    IV. 


PROPOSITION  XIII.    PROBLEM. 

383,  To  construct  a  square  equivalent  to  the  sum 
of  any  number  of  given  squares. 


K 


0 
p 


Let  m,  n,  o,  p,  r  be  sides  of  the  given  squares. 

To  construct  a  square  =c=  m2  +  ri*  +  o2  +p2  +  r2. 
Construction,  Take  AB  =  m. 

Draw  AC  =  n  and  J.  to  AB  at  A,  and  draw  EC. 

Draw  CE  =  o  and  J_  to  BC  at  C,  and  draw  BE. 

Draw  EF  =  p  and  _L  to  BE  at  E,  and  draw  BF. 

Draw  FH=  r  and  ±  to  BF  at  .F,  and  draw  BH. 
The  square  constructed  on  BH\&  the  square  required. 

Proof, 


-  FH2  +  W+EF*  +  CA2  +  A£\      §  379 

uares  on  the  two  leas  of  a  rt.  A  is  equiv 
on  the  hypotenuse). 

That  is,          BlF  **  m2  +  n2  +  o2  +p  +  r2, 


(the  sum  of  the  squares  on  the  two  leas  of  a  rt.  A  is  equivalent  to  the  square 
on  the  hypotenuse). 


Q.E.F. 


PROBLEMS   OF   CONSTRUCTION. 


195 


PROPOSITION  XIV.     PROBLEM. 

384,  To  construct  a  polygon  similar  to  two  given 
similar  polygons  and  equivalent  to  their  sum. 


P  U 

Let  R  and  R'  be  two  similar  polygons,  and  AB  and 
A'B'  two  homologous  sides. 

To  construct  a  similar  polygon  equivalent  to  R  +  -Rf. 
Construction,          Construct  the  rt.  Z  P. 

Take  PH=  A'B',  and  PO  =  AB. 

Draw  OH,  and  take  A"£"  =  OH. 

Upon  A"I>f',  homologous  to  AB,  construct  R"  similar  to  R. 
Then  R"  is  the  polygon  required. 

Proof,   PO2  +  PII*  =  OH\  .'.  AB*  +  A^B^  =  A"£"\ 


Now 


and 


§376 


(similar  polygons  are  to  each  other  as  the  squares  of  their  homologous  sides). 


By  addition,    ^  +  ^  = 


19G 


PLANE   GEOMETEY.  -  BOOK    IV. 


PROPOSITION  XV.     PROBLEM. 

385,  To  construct  a  polygon  similar  to  two  given 
similar  polygons  and  equivalent  to  their  difference. 


A!  B'  A.  13  A'         B"        P  O 

Let  R  and  Rr  be  two  similar  polygons,  and  AB  and 
A'B'  two  homologous  sides. 

To  construct  a  similar  polygon  equivalent  to  M1  —  H. 
Construction,          Construct  the  rt.  Z.  P, 
and  take  PO  =  AB. 

From  0  as  a  centre,  with  a  radius  equal  to  A1!?, 
describe  an  arc  cutting  PX  at  H,  and  join  OH. 
Take  A"J3"  =  PIT,  and  on  A"£",  homologous  to 

construct  R"  similar  to  R. 
Then  JR1t  is  the  polygon  required. 

Proof. 


Now 


and 


§376 


(similar  polygons  are  to  each  other  as  the  squares  of  their  homologous  sides). 
By  subtraction, 


E" 


PROBLEMS  OF  CONSTRUCTION. 


197 


PEOPOSITION  XVI.     PROBLEM. 

386,   To  construct  a  triangle  equivalent  to  a  given 
polygon. 


E 


F 


Let  ABCDHE  be  the  given  polygon. 

To  construct  a  triangle  equivalent  to  the  given  polygon. 
Construction,  From  D  draw  DE, 

and  from  J7  draw  HF  \\  to  DE. 
Produce  AE  to  meet  HF  &i  F,  and  draw  DF. 
Again,  draw  OF,  and  draw  DK  II  to  OF  to  meet  AF  pro- 
duced at  K,  and  draw  CK. 

In  like  manner  continue  to  reduce  the  number  of  sides  of 
the  polygon  until  we  obtain  the  A  CIK. 

Proof,    The  polygon  ABCDF  has  one  side  less  than  the 
polygon  ABCDHE,  but  the  two  are  equivalent. 
For  the  part  ABODE  is  common, 

and  the  A  DEF^  A  DEE,  §  369 

(for  the  base  DE  is  common,  and  their  vertices  F  and  H  are  in  the  line 
FH II  to  the  base). 

The  polygon  ABCK  has  one  side  less  than  the  polygon 
ABCDF,  but  the  two  are  equivalent. 

For  the  part  AJBCFis  common, 

and  the  A  CFK**  A  CFD,  §  369 

(for  the  base  CF  is  common,  and  their  vertices  K  and  D  are  in  the  line 
KD  II  to  the  base\ 

In  like  manner  the  A  CIK**  ABCK. 

Q  E.  F. 


198  PLANE   GEOMETRY.  —  BOOK   IV. 


PROPOSITION  XVII.     PROBLEM. 

387,   To  construct  a  square  which  shall  have  a  given 

ratio  to  a  given  square* 

D 


/    /    3 

/    / 

^/      a 
m 


n. 


£T^-,      / 

n*-^j 


Let  R  be  the  given  square,  and  -  the  given  ratio. 
_  >  m 

To  construct  a  square  which  shall  be  to  R  as  n  is  to  m. 

Construction,  Take  AB  equal  to  a  side  of  R,  and  draw  Ay, 
making  any  acute  angle  with  AB. 

On  Ay  take  AU=m,  EF=  n,  and  join  EB. 
Draw  FC  \\  to  EB  to  meet  AB  produced  at  C. 

On  AC  as  a  diameter  describe  a  semicircle. 
At  B  erect  the  _L  BD,  meeting  the  semi  circumference  at  D. 

Then  BD  is  a  side  of  the  square  required. 
Proof,    Denote  AB  by  a,  BC\>y  b,  and  BD  by  x. 

Now  a  :  x  =  x  :  b  ;  that  is,  x2  =  ab.  §  337 

Hence,  a2  will  have  the  same  ratio  to  z?  and  to  ab. 
Therefore  a*  :  x2  =  a2  :  ab  —  a  :  b. 

:b  =  m:n,  §  309 


(a  straight  line  drawn  through  two  sides  of  a  A,  parallel  to  the  third  side, 
divides  those  sides  proportionally). 

Therefore  a2  :  x2  —  m  :  n. 
By  inversion,  x2  :  a2  =  n  :  on. 

Hence  the  square  on  BD  will  have  the  same  ratio  to  E  as 
n  has  to  m.  o.  E.  F, 


PROBLEMS   OF   CONSTRUCTION.  199 


PROPOSITION  XVIII.    PROBLEM. 

388,   To  construct  a  polygon  similar  to  a  given  poly- 
gon and  having  a  given  ratio  to  it. 


— 

v i 

Let  R  be  the  given  polygon  and  —  the  given  ratio. 

To  construct  a  polygon  similar  to  R,  which  shall  be  to  R  as 
n  is  to  m. 

Construction,  Find  a  line  A'B1,  such  that  the  square  con- 
structed upon  it  shall  be  to  the  square  constructed  upon  AE 
*is  n  is  to  m.  §  387 

Upon  A1!!'  as  a  side  homologous  to  AB,  construct  the  poly- 
gon /S  similar  to  R. 

Then  S  is  the  polygon  required. 

Proof,  8\R  =  AB1' :  AB\  §  376 

(similar  polygons  are  to  each  other  as  the  squares  of  their  homologous  sides). 
But  J72?'2 :  ABZ  =  n:m.  Cons. 

«  Therefore  S :  R  —  n  :  m. 

Q.  E.  F. 


Ex.  304.  Find  the  area  of  a  right  triangle  if  the  length  of  the  hypote- 
nuse is  17  feet,  and  the  length  of  one  leg  is  8  feet. 

Ex.  305.  Compare  the  altitudes  of  two  equivalent  triangles,  if  the 
base  of  one  is  three  times  that  of  the  other. 

Ex,  306.  The  bases  of  a  trapezoid  are  8  feet  and  10  feet,  and  the  alti- 
tude is  6  feet.  Find  the  base  of  an  equivalent  rectangle  having  an  equal 
altitude. 


200 


PLANE   GEOMETRY.  —  BOOK   IV. 


PKOPOSITION  XIX.     PROBLEM. 

389.  To  construct  a  square  equivalent  to  a  given 
parallelogram. 


p 


R 


l ,   Ml 


\ 


N  O 

Let  ABCD  be  a  parallelogram,  b  its  base,  and  a  its 
altitude. 

To  construct  a  square  equivalent  to  the  EH  AS  CD. 

Construction,    Upon  the  line  JfJf  take  MN~  a,  and  N0  =  b. 

Upon  MO  as  a  diameter,  describe  a  semicircle. 
At  N  erect  NP  J_  to  MO,  to  meet  the  circumference  at  P. 

Then  the  square  R,  constructed  upon  a  line  equal  to  NP, 
is  equivalent  to  the  £7  ABCD. 

Proof,  MN :  NP  =  NP  :  NO,  §  337 

(a  _L  let  fall  from  any  point  of  a  circumference  to  the  diameter  is  a  mean 
proportional  between  the  segments  of  the  diameter). 

.'.  NP*  =  MNx  N0  =  axb. 
That  is,  R^OABCD. 

Q.  E.  F. 

• 

390,  Con.  1.  A  square  may  be  constructed  equivalent  to  a 
given  triangle,  by  taking  for  its  side  a  mean  proportional  be- 
tween the  base  and  one-half  the  altitude  of  the  triangle. 

391,  COR.  2.  A  sqiicwe  may  be  constructed  equivalent  to  a 
given  polygon,  by  first  reducing  the  polygon  to  an  equivalent 
triangle,  and  then   constructing  a  square  equivalent   to   the 
triangle 


PROBLEMS   OF   CONSTRUCTION.  201 


PROPOSITION  XX.     PROBLEM. 

392,  To  construct  a  parallelogram  equivalent  to  a 
given  square,  and,  having  the  sum  of  its  base  and 
altitude  equal  to  a  given  line. 


R 


Let  R  be  the  given  square,  and  let  the  sum  of  the 
base  and  altitude  of  the  required  parallelogram  be 
equal  to  the  given  line  MN. 

To  construct  a  £7  equivalent  to  R,  with  the  sum  of  its  base 
and  altitude  equal  to  MN. 

Construction,  Upon  JOT  as  a  diameter,  describe  a  semicircle. 
At  M  erect  a  JL  MP,  equal  to  a  side  of  the  given  square  fi. 
Draw  PQ  II  to  MN,  cutting  the  circumference  at  8. 
Draw  SC±  to  MN. 

Any.O  having  CM  for  its  altitude  and  CTVfor  its  base  is 
equivalent  to  R. 

Proof,  80=  PM.  §§  100,  180 

.'.  SO*  =  PM*  =  R. 

But  MO:  80  =80:  ON,  §  337 

(a  J_  let  fall  from  any  point  in  the  circumference  io  the  diameter  is  a  mean 
proportional  between  the  segments  of  the  diameter). 

Then  SC'^MOx  ON.  Q.  E.  F. 

NOTE.  This  problem  may  be  stated :  To  construct  two  straight  tinea 
the  sum  and  product  of  which  are  known. 


202  PLANE    GEOMETRY. BOOK    IV. 


PROPOSITION  XXI.     PROBLEM. 

393,  To  construct  a  parallelogram  equivalent  to  a 
given  square,  and  having  the  difference  of  its  base 
and  altitude  equal  to  a  given  line. 


R' 


Let  R  be  the  given  square,  and  let  the  difference  of 
the  base  and  altitude  of  the  required  parallelogram 
be  equal  to  the  given  line  MN. 

To  construct  a  O  equivalent  to  R,  with  the  difference  of  the 
base  and  altitude  equal  to  MN. 

Construction.  Upon  the  given  line  JJOTas  a  diameter,  describe 
a  circle. 

From  M  draw  MS,  tangent  to  the  O,  and  equal  to  a  side 
of  the  given  square  R. 

Through  the  centre  of  the  O  draw  SB  intersecting  the  cir- 
cumference at  C  and  B. 

Then  any  O,  as  R' ,  having  SB  for  its  base  and  SO  for  its 
altitude,  is  equivalent  to  R. 

Proof,  SB  :  SM=  SM :  SO,  §  348 

(if from  a  point  without  aQa  secant  and  a  tangent  are  drawn,  the  tangent  is 
a  mean  proportional  between  the  whole  secant  and  the  part  without  the  O). 

Then  SM*-SBxSC, 

and  the  -difference  between  SB  and  SO  is  the  diameter  of  the 
O,  that  is,  JOT.  Q.E.F. 

NOTE.  This  problem  may  be  stated  :  To  construct  two  straight  lines 
the  difference  and  product  of  which  are  known. 


PROBLEMS   OF   CONSTRUCTION.  203 


PROPOSITION  XXII.     PROBLEM. 

394,  To  construct  a  polygon  similar  to  a  given  poly- 
gon P,  and  equivalent  to  a  given  polygon  Q. 


P' 

\ 


n,^     /  /  m 

___       _j_  __ 

Let  P  and  Q  be  two  polygons,  and  AB  a  side  of  P. 

To  construct  a  polygon  similar  to  P  and  equivalent  to  Q. 
Construction,    Find  squares  equivalent  to  P  and  Q,       §  391 

and  let  m  and  n  respectively  denote  their  sides. 
Find  A'B1,  a  fourth  proportional  to  m,  n,  and  AB.      §  351 
Upon  A'B1,  homologous  to  AB,  construct  P  similar  to  P. 

Then  P'  is  the  polygon  required. 
Proof.  m  :  n  =  AB  :  A'B',  Cons. 

.'.  m2 :  n2  =  AB  :  A'B' . 

But  P=c=m2,  and  Q=s=n2.  Cons. 

.:P:Q  =  m*:n*  =  AB* :  A'B1'. 

But  P:Pt=--AB2:ArB'\  §376 

(similar  polygons  are  to  each  other  as  the  squares  of  their  homologous  sides). 

.•.P\Q=P:P.  Ax.  1 

/.  P1  is  equivalent  to  Q,  and  is  similar  to  P  by  construction. 

Q.  E.  F. 


204 


PLANE   GEOMETRY.  —  BOOK   IV. 


PROBLEMS  OF  COMPUTATION. 

Ex.  307.    To  find  the  area  of  an  equilateral  triangle  in  terms  of  its 
side. 

Denote  the  side  by  a,  the  altitude  by  h,  and  the  area  by  S. 

Then  W  =  a2  -  -  =  ^- 


But 


axh 


Ex.  308.   To  find  the  area  of  a  triangle  in  terms  of  its  sides. 


By  Ex.  219, 
Hence, 


h  =  Y  Vs  (s  —  a)  (s  —  b)  (s  —  c). 
o 

S=  -  X  j-  Vs(s-a)(s-b)(s-c) 
=  Vs  (s  —  a)  (s  -  b)  (s  —  c). 


Ex.  309.   To  find  the  area  of  a  triangle  in  terms  of  the  radius  of  the 
circumscribing  circle. 

If  R  denote  the  radius  of  the  circumscribing  circle,  alid  h  the  altitude 
of  the  triangle,  we  have,  by  Ex.  222,  4jt 


Multiply  by  a,  and  we  have 

But  axh  =  28. 

'  CL  x  b  x  c  =  4  R  x  S 
o     abc 


NOTE.  The  radius  of  the  circumscribing  circle  is  equal  to 


abc 


EXEECISES.  205 


THEOREMS. 

310.  In  a  right  triangle  the  product  of  the  legs  is  equal  to  the  product 
of  the  hypotenuse  and  the  perpendicular  drawn  to  the  hypotenuse  from 
the  vertex  of  the  right  angle. 

311.  If  ABO  is  a  right  triangle,  0  the  vertex  of  the  right  angle, 
BD  a  line  cutting  AC  in  D,  then  BD*  +  AC2  -  AB2  +  TX?. 

312.  Upon  the  sides  of  a  right  triangle  as  homologous  sides  three 
similar  polygons  are  constructed.     Prove  that  the  polygon  upon   the 
hypotenuse  is  equivalent  to  the  sum  of  the  polygons  upon  the  legs. 

313.  Two  isosceles  triangles  are  equivalent  if  their  legs  are  equal  each 
to  each,  and  the  altitude  of  one  is  equal  to  half  the  base  of  the  other. 

/C  31C   The  area  of  a  circumscribed  polygon  is  equal  to  half  the  product 
of  its  perimeter  by  the  radius  of  the  inscribed  circle. 

315.  Two  parallelograms  are  equaJfi4,$S^  adjacent  sides  of  the  one 
are  equal  respectively  to  two  adjacent  skMtof  the  other,  and  the  included 
angles  are  supplementary.  "^^j^ON^ 

316.  Every  straight  line  drawn  through  the, centre  of  a  parallelogram 
divides  it  into  two  equal  parts. 

317.  If  the  middle  points  of  two  adjacent  sides  of  a  parallelogram  are 
joined,  a  triangle  is  formed  which  is  equivalent  to  one-eighth  of  the 
entire  parallelogram. 

318.  If  any  point  within  a  parallelogram  is  joined  to  the  four  vertices, 
the  sum  of  either  pair  of  triangles  having  parallel  bases  is  equivalent  to 
one-half  the  parallelogram. 

319.  The  line  which  joins  the  middle  points  of  the  bases  of  a  trape- 
zoid divides  the  trapezoid  into  two  equivalent  parts. 

-  320.   The  area  of  a  trapezoid  is  equal  to  the  product  of  one  of  the  legs 
and  the  distance  from  this  leg  to  the  middle  point  of  the  other  leg. 

321.  The  lines  joining  the  middle  point  of  the  diagonal  of  a  quadri- 
lateral to  the  opposite  vertices  divide  the  quadrilateral  into  two  equiva- 
lent parts. 

N,  322.    The  figure  whose  vertices  are  the  middle  points  of  the  sides  of 
any  quadrilateral  is  equivalent  to  one-half  of  the  quadrilateral. 

323.  ABO  is  a  triangle,  M  the  middle  point  of  AB,  P  any  point  in 
AB  between  A  and  M.  If  MD  is  drawn  parallel  to  PC,  and  meeting 
BO  at  D,  the  triangle  BPD  is  equivalent  to  one-half  the  triangle  ABC. 


206  PLANE   GEOMETRY.  —  BOOK   IV. 


NUMERICAL  EXERCISES. 

324.  Find  the  area  of  a  rhombus,  if  the  sum  of  its  diagonals  is  12  feet, 
and  their  ratio  is  3  :  5.          g \*    ^?~)  y* 

325.  Find  the  area  of  an  isosceles  right  triangle  if  the  hypotenuse 
is  20  feet         /  fr  t 

326.  In  a  right  triangle,  the  hypotenuse  is  13  feet,  one  leg  is  5  feet. 
Find  the  area.    • 

327.  Find  the  area  of  an  isosceles  triangle  if  the  hase  =  &,  and  leg  =  c. 

328.  Find  the  area  of  an  equilateral  triangle  if  one  side  =  8.  /C  U 

329.  Find  the  area  of  an  equilateral  triangle  if  the  altitude  =  h. 

330.  A  house  is  40  feet  long,  30  feet  wide,  25  feet  high  to  the  eaves, 
and  35  feet  high  to  the  ridge-pole.     Find  the  number  of  squarefeet  in  fy 
its  entire  exterior  surface.  ^f 

331.  The  sides  of  a  right  triangle  are  as  3  :  4  :  5.     The  altrrcrae  upon 
the  hypotenuse  is  12  feet.     Find  the  area. 

332.  Find  the  area  of  a  right  triangle  if  one  leg  =  a,  and  the  altitude 
upon  the  hypotenuse  =  h. 

333.  Find  the  area  of  a  triangle  if  the  lengths  of  the  sides  are  104 
feet,  111  feet,  and  175  feet. 

r    334.   The. area  of  a  trapezoid  is  700  square  feet.    The  bases  are  30  feet 
and  40  feet  respectively.     Find  the  distance  between  the  bases. 

335.  ABCD  is  a  trapezium;  ^4J5  =  87feet,  B0=  119  feet,  CD  =  41 
feet,  DA  =  169  feet,  AC=  200  feet.     Find  the  area. 

336.  What  is  the  area  of  a  quadrilateral  circumscribed  about  a  circle 
whose  radius  is  25  feet,  if  the  perimeter  of  the  quadrilateral  is  400  feet? 
What  is  the  area  of  a  hexagon  having  an  equal  perimeter  and  circum- 
scribed about  the  same  circle  ? 

337.  The  base  of  a  triangle  is  15  feet,  and  its  altitude  is  8  feet.    Find 
the  perimeter  of  an  equivalent  rhombus  if  the  altitude  is  6  feet.     /  C 

338.  Upon  the  diagonal  of  a  rectangle  24  feet  by  10  feet  a  triangle 
equivalent  to  the  rectangle  is  constructed.     What  is  its  altitude  ? 

339.  Find  the  side  of  a  square  equivalent  to  a  trapezoid  whose  bases 
are  56  feet  and  44  feet,  and  each  leg  is  10  feet.     |  (f  •  j 

340.  Through  a  point  Pin  the  side  AB  of  a  triangle  ABC,  a  line  is 
drawn  parallel  to  BC,  and  so  as  to  divide  the  triangle  into  two  equiva- 
lent parts.     Find  the  value  of  A P  in  terms  of  AB. 


EXERCISES.  207 

341.  What  part  of  a  parallelogram  is  the  triangle  cut  off  by  a  line 
drawn  from  one  vertex  to  the  middle  point  of  one  of  the  opposite  sides  ? 

342.  In  two  similar  polygons,  two  homologous  sides  are  15  feet  and 
25  feet.     The  area  of  the  first  polygon  is  450  square  feet.    Find  the  area 
of  the  other  polygon. 

343.  The  base  of  a  triangle  is  32  feet,  its  altitude  20  feet.     What  is 
the  area  of  the  triangle  cut  off  by  drawing  a  line  parallel  to  the  base 
and  at  a  distance  of  15  feet  from  the  base  ? 

344.  The  sides  of  two  equilateral  triangles  are  3  feet  and  4  feet.    Find 
the  side  of  an  equilateral  triangle  equivalent  to  their  sum. 

345.  If  the  side  of  one  equilateral  triangle  is  equal  to  the  altitude  of 
another,  what  is  the  ratio  of  their  areas  ? 

346.  The  sides  of  a  triangle  are  10  feet,  17  feet,  and  21  feet.     Find 
the  alfcfi  of  the  parts*  in  to  which  the  triangle  is  divided  by  bisecting  the 
angle^^led  by  the  first  two  sides. 

347.  In  a  trap^oid,  one  base  is  10  feet,  the  altitude  is  4  feet,  the  area 
is  32  square  feet.v  Find  the  length  of  a  line  drawn  between  the  legs 
parallel  to  the  base  and  distant  1  foot  from  it. 

348.  If  the  altitude  h  of  a  triangle  is  increased  by  a  length  m,  how 
much  must  be  taken  from  the  base  a  in  order  that  the  area  may  remain 
the  same  ? 

349.  Find  the  area  of  a  right  triangle,  having  given  the  segments  p, 
q,  into  which  the  hypotenuse  is  divided  by  a  perpendicular  drawn  to  the 
hypotenuse  from  the  vertex  of  the  right  angle. 

PROBLEMS. 

850.   To  construct  a  triangle  equivalent  to  a  given  triangle,  and 
having  one  side  equal  to  a  given. length  I. 

351.  To  transform  a  triangle  into  an  equivalent  right  triangle. 

352.  To  transform  a  triangle  into  an  equivalent  isosceles  triangle. 

353.  To  transform  a  triangle  ABO  into  an  equivalent  triangle,  hav- 
ing one  side  equal  to  a  given  length  I,  and  one  angle  equal  to  angle  BAG. 

HINTS.   Upon  AB  (produced  if  necessary),  take  AD  =  I,  draw  BE  II  to 
CD,  and  meeting  AC  (produced  if  necessary)  at  E\  A  BED^k  BEO. 

354.  To  transform  a  given  triangle  into  an  equivalent  right  triangle, 
having  one  leg  equal  to  a  given  length. 


. 

208  PLANE   GEOMETRY.  —  BOOK   IV. 

355.  To  transform  a  given  triangle  into  an  equivalent  right  triangle, 
having  the  hypotenuse  equal  to  a  given  length. 

356.  To  transform  a  given  triangle  into  an  equivalent  isosceles  tri- 
angle, having  the  base  equal  to  a  given  length. 

To  construct  a  triangle  equivalent  to: 

357.  The  sum  of  two  given  triangles. 

358.  The  difference  of  two  given  triangles. 

359.  To  transform  a  given  triangle  into  an  equivalent  equilateral 
triangle. 

To  transform  a  parallelogram  into : 

360.  A  parallelogram  having  one  side  equal  to  a  given  length. 

361.  A  parallelogram  having  one  angle  equal  to  a  given  angle. 

362.  A  rectangle  having  a  given  altitude. 
To  transform  a  square  into  : 

363.  An  equilateral  triangle. 

364.  A  right  triangle  having  one  leg  equal  to  a  given  length. 

365.  A  rectangle  having  one  side  equal  to  a  given  length. 

To  construct  a  square  equivalent  to : 

366.  Five-eighths  of  a  given  square. 
>867.   Three-fifths  of  a  given  pentagon. 

'368.  To  draw  a  line  through  the  vertex  of  a  given  triangle  so. as  to 
divide  the  triangle  into  two  triangles  which  shall  be  to  each  other  as  2 :  3. 
*s  369.  To  divide  a  given  triangle  into  two  equivalent  parts  by  drawing 
a  line  through  a  given  point  P  in  one  of  the  sides. 

370.    To  find  a  point  within  a  triangle,  such  that  the  lines  joining  this 
point  to  the  vertices  shall  divide  the  triangle  into  three  equivalent  parts. 

*371.  To  divide  a  given  triangle  into  two  equivalent  parts  by  drawing 
a  line  parallel  to  one  of  the  sides. 

372.  To  divide  a  given  triangle  into  two  equivalent  parts  by  drawing 
a  line  perpendicular  to  one  of  the  sides. 

^373.   To  divide  a  given  parallelogram  into  two  equivalent  parts  by 
drawing  a  line  through  a  given  point  in  one  of  the  sides. 

374.  To  divide  a  given  trapezoid  into  two  equivalent  parts  by  draw- 
ing a  line  parallel  to  the  bases. 

^B75.  To  divide  a  given  tranezoid  into  two  equivalent  parts  by  draw- 
ing a  line  through  a  given  point  in  one  of  the  bases. 


BOOK  V. 
REGULAR    POLYGONS   AND   CIRCLES. 

395,  A  regular  polygon  is  a  polygon  which  is  equilateral 
and  equiangular ;  as,  for  example,  the  equilateral  triangle,  and 
the  square. 

PROPOSITION  I.     THEOREM. 

396,  An  equilateral  polygon  inscribed  in  a  circle  is 
a  regular  polygon. 


Let  ABC  etc.,  be  an  equilateral  polygon  inscribed  in 
a,  circle. 

To  prove  the  polygon  ABC  etc.,  regular. 

Proof,        The  arcs  AB,  BC,  CD,  etc.,  are  equal,  §  230 

(in  the  same  O,  equal  chords  subtend  equal  arcs). 

Hence  arcs  ABC,  BCD,  etc.,  are  equal,  Ax.  6 

and  the  A  A,  B,  C,  etc.,  are  equal, 

(being  inscribed  in  equal  segments). 

Therefore  the  polygon  ABC,  etc.,  is  a  regular  polygon,  being 
equilateral  and  equiangular.  Q.  E.  D. 


210 


PLANE   GEOMETRY.  —  BOOK    V. 


PROPOSITION  II.     THEOREM. 

397,  A  circle  may  be   circumscribed   about,  and  a 
circle  may  be  inscribed  in,  any  regular  polygon. 

D 


Let  ABODE  be  a  regular  polygon. 

I.    To   prove    that   a   circle   may   be   circumscribed  about 
ABODE, 

Proof,    Let.  0  be  the  centre  of  the  circle  passing  through 
A,  £,  0. 

Join  OA,  OB,  00,  and  OD. 

Since  the  polygon  is  equiangular,  and  the  A  OBCis  isosceles,  - 


and  ZOBC  =  ./OCB.  §  154 

By  subtraction,        Z  OB  A  =  Z  OCD. 
Hence  in  the  A  OB  A  and  OCD 

the  Z  OB  A  =  Z  OCD, 
the  radius  OB  —  the  radius  OC, 
and  AB  --=  CD.  §  395 

.-.AOAB  =  AOCD,  §150 

(having  two  sides  and  the  included  /.  of  the  one  equal  to  two  sides  and  the 
included  Z.  of  the  other). 

.-.  OA  =  OD. 

Therefore  the  circle  passing  through  A,  J3,  and  (7,  also 
passes  through  D. 


REGULAR   POLYGONS   AND   CIRCLES.  211 

In  like  manner  it  may  be  proved  that  the  circle  passing 
through  B,  C,  and  Dt  also  passes  through  E\  and  so  on 
through  all  the  vertices  in  succession. 

Therefore  a  circle  described  from  0  as  a  centre,  and  with  a 
radius  OA,  will  be  circumscribed  about  the  polygon. 

II.    To  prove  that  a  circle  may  be  inscribed  in  ABODE. 

Proof,  Since  the  sides  of  the  regular  polygon  are  equal 
chords  of  the  circumscribed  circle,  they  are  equally  distant 
from  the  centre.  §  236 

Therefore  a  circle  described  from  0  as  a  centre,  and  with 
the  distance  from  0  to  a  side  of  the  polygon  as  a  radius,  will 
be  inscribed  in  the  polygon.  Q.  E.D. 

398,  The  radius  of  the  circumscribed  circle,  OA,  is  called 
the  radius  of  the  polygon. 

399,  The  radius  of  the  inscribed  circle,  OF,  is  called  the 
apothem  of  the  polygon. 

400,  The  common  centre  0  of  the  circumscribed  and  in- 
scribed circles  is  called  the  centre  of  the  polygon. 

401,  The  angle  between  radii  drawn  to  the  extremities  of 
any  side,  as  angle  AOB,  is  called  the  angle  at  the  centre  of  the 
polygon. 

By  joining  the  centre  to  the  vertices  of  a  regular  polygon, 
the  polygon  can  be  decomposed  into  as  many  equal  isosceles 
triangles  as  it  has  sides.  Therefore, 

402,  COR.  1.    The  angle  at  the  centre  of  a  regular  polygon  is 
equal  to  four  right  angles  divided  by  the  number  of  sides  of 
the  polygon. 

403,  COR.  2.    The  radius  drawn  to  any  vertex  of  a  regular 
polygon  bisects  the  angle  at  the  vertex. 

404,  COR.  3.    The  interior  angle  of  a  regular  polygon  is  the 
supplement  of  the  angle  at  the  centre. 

For  the  Z  ABO  =  2  Z  ABO  =  Z  ABO  +  Z.BAO.  Hence 
the  Z  ABC  is  the  supplement  of  the  Z  A0£. 


212  PLANE    GEOMETRY. BOOK   V. 

PROPOSITION  III.     THEOREM. 

405,  If  the  circumference  of  a  circle  is  divided  into 
any  number  of  equal  parts,  the  chords  joining  the 
successive  points  of  division  form  a  regular  inscribed 
polygon,  and  the  tangents  drawn  at  the  points  of 
division  form  a  regular  circumscribed  polygon. 
I  D  H 


Let  the  circumference  be  divided  into  equal  arcs, 
AB,  BC,  CD,  etc.,  be  chords,  FBG,  GCH,  etc.,  be  tangents. 

I.  To  prove  that  ABODE  is  a  regular  polygon. 

Proof,       The  sides  AB,  BC,  CD,  etc.,  are  equal,          §  230 

(in  the  same  O  equal  arcs  are  subtended  by  equal  chords). 

Therefore  the  polygon  is  regular,  §  396 

(an  equilateral  polygon  inscribed  in  a  0  is  regular). 

II.  To  prove  that  the  polygon  FGHIK  is  a  regular  polygon. 
Proof,    In  the  A  AFB,  BGC,  CUD,  etc. 

A3  =  £0=  CD,  etc.  §395 

Also,  Z  BAF=  Z  ABF=  Z.  CBG  =  Z  BOO,  etc.,    §  269 

(being  measured  by  halves  of  equal  arcs). 
Therefore  the  triangles  are  all  equal  isosceles  triangles. 

Hence  Z.F  =  Z.G  =  Z.H,  etc. 
Also,  FB^BG^GC  =  CH,  etc. 

Therefore  FQ  =  GIT,  etc. 
/.  FGHIK  is  a  regular  polygon.  §  395 

Q.  E.  D. 

406,  COR.  1.  Tangents  to  a  circumference  at  the  vertices  of  a 
regular  inscribed  polygon  form  a  regular  circumscribed  poly- 
gon of  the  same  number  of  sides. 


REGULAR   POLYGONS   AND   CIRCLES. 


213 


407,  COR.  2.    If  a  regular  polygon  is  inscribed  in  a  circle, 
the  tangents  drawn  at  the  middle  points 

of  the  arcs  subtended  by  the  sides  of  the 
polygon  form  a  circumscribed  regular 
polygon,  whose  sides  are  parallel  to  the 
sides  of  the  inscribed  polygon  and  whose 
vertices  lie  on  the  radii  (prolonged}  of 
the  inscribed  polygon.  For  any  two  cor- 
responding sides,  as  AB  and  A'B',  perpendicular  to  OM, 
are  parallel,  and  the  tangents  MB'  and  NB1 ,  intersecting  at  a 
point  equidistant  from  OJfand  OJV(§  246),  intersect  upon  the 
bisector  of  the  Z  MON(§  163)  ;  that  is,  upon  the  radius  OB. 

408,  COR.  3.  If  the  vertices  of  a  regular  inscribed  polygon 
are  joined  to  the  middle  points  of  the  arcs  sub- 
tended by  the  sides  of  the  polygon,  the  joining 

lines  form   a  regular   inscribed  polygon    of 
double  the  number  of  sides. 

409,  COR.  4.  Jf  tangents  are  drawn  at  the 
middle  points   of  the  arcs   between  adjacent 
points  of  contact  of  the  sides  of  a  regular  cir- 
cumscribed polygon,  a  regular  circumscribed 
polygon  of  double   the    number  of   sides  is 

f°rmed'  A     E        *      B 

410,  SCHOLIUM.    The  perimeter  of  an  inscribed  polygon  is 
less  than  the  perimeter  of  the  inscribed  polygon  of  double  the 
number  of  sides;  for  each  pair  of  sides  of  the  second  polygon 
is  greater  than  the  side  of  the  first  polygon  which  they  replace 
(§  137). 

The  perimeter  of  a  circumscribed  polygon  is  greater  than 
the  perimeter  of  the  circumscribed  polygon  of  double  the  num- 
ber of  sides ;  for  every  alternate  side  FG,  HI,  etc.,  of  the  poly- 
gon FGHI,  etc.,  replaces  portions  of  two  sides  of  the  circum- 
scribed polygon  ABCD,  and  forms  with  them  a  triangle,  and 
one  side  of  a  triangle  is  less  than  the  sum  of  the  other  two  sides. 


214 


PLANE   GEOMETRY.  —  BOOK   V. 


PROPOSITION  IV.    THEOREM. 

411,  Two  regular  polygons  of  the  same  number  of 
sides  are  similar. 


A  B  A'  B' 

Let  Q  and  Q'  be  two  regular  polygons,  each  having 
n  sides. 

To  prove  Q  and  Q1  similar  polygons. 

Proof,  The  sum  of  the  interior  A  of  each  polygon  is  equal  to 

(n  -  2)  2  rt.  A,  §  205 

(the  sum  of  the  interior  A  of  a  polygon  is  equal  to  2  rt.  A  taken  as  many 
times  less  2  as  the  polygon  has  sides). 

Each  angle  of  either  polygon  —  -  -  *—  —  :  —  >     §  206 

n 

(for  the  A  of  a  regular  polygon  are  all  equal,  and  hence  each  Z.  is  equal 
to  the  sum  of  the  A  divided  by  their  number). 

Hence  the  two  polygons  Q  and  Q'  are  mutually  equiangular. 
Since  AB  =  EC,  etc.,  and  A'ff  =  B'C\  etc.,          §  395 


Hence  the  two  polygons  have  their  homologous  sides 
proportional. 

Therefore  the  two  polygons  are  similar.  §  319 


Q.E.  D. 


412,  COR.  The  areas  of  two  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  the  squares  of  any  two 
homologous  sides,  §  376 


REGULAR   POLYGONS   AND   CIRCLES. 


215 


PROPOSITION  V.     THEOREM. 

413,  The  perimeters  of  two  regular  polygons  of  the 
same  number  of  sides  are  to  each  other  as  the  radii 
of  their  circumscribed  circles,  and  also  as  the  radii 
of  their  inscribed  circles. 

D' 

D 


A     M    B  A'      M'     B' 

Let  P  and  P'  denote  the  perimeters,  0  and  0>  the 
centres,  of  the  two  regular  polygons. 

From  0,  0'  draw  OA,  O'A',  OB,  0'£',  and  Js  OM,  O'M1. 
To  prove       P :  P  =  OA  :  O'A'  =  OM:  O'M1. 
Proof,   Since  the  polygons  are  similar,  §  411 

§333 


In  the  isosceles  A  GAB  and  O'A'B' 

theZO-theZO', 
and  OA  :  OB  =  O'A'  :  0'3'. 
.'.  the  A  GAB  and  O'A'ff  are  similar. 


§402 


§326 
§319 

Also  AB  :  A'J?  =  OM:  O'M1,  §  328 

(the  homologous  altitudes  of  similar  &  have  the  same  ratio  as  their  bases). 
.'.  P:P'=OA:  O'A'  =  OM:  OM'. 

Q.  E.  D. 

414,  COR.  The  areas  of  two  regular  polygons  of  the  same 
number  of  sides  are  to  each  other  as  the  squares  of  the  radii 
of  their  circumscribed  circles,  and  also  as  the  squares  of  the 
radii  of  their  inscribed  circles.  §  376 


216 


PLANE   GEOMETRY.  —  BOOK   V. 


PROPOSITION  VI.     THEOREM. 

415.  The  difference  between  the  lengths  of  the  perim- 
eters of  a  regular  inscribed  polygon  and  of  a  similar 
circumscribed  polygon  is  indefinitely  diminished  as 
the  number  of  the  sides  of  the  polygons  is  indefinitely 
increased. 


§333 

§  301 
§  295 


Let  P  and  Pf  denote  the  lengths  of  the  perimeters, 
AB  and  A'B*  two  corresponding  sides,  OA  and  OA1  the 
radii,  of  the  polygons. 

To  prove  that  as  the  number  of  the  sides  of  the  polygons  is 
indefinitely  increased,  P  —  P  is  indefinitely  diminished. 

Proof,    Since  the  polygons  are  similar, 
P:P=OA':OA. 

Therefore  P  —  P  :  P  : :  OA'  —  OA  :  OA. 

Whence  OA(P  —  P)  =  P(OA'  —  OA). 

Now  OA  is  the  radius  of  the  circle,  and  P,  though  an 
increasing  variable,  always  remains  less  than  the  circumference 
of  the  circle. 

Therefore  P  —  P  is  indefinitely  diminished,  if  OA1 —  OA 
is  indefinitely  diminished. 

Draw  the  radius  OO  to  the  point  of  contact  of  A'H1. 

In  the  A  OA'C,       OA'  -  OC<  AC.  §  137 

Substituting  OA  for  its  equal  00,  we  have 
OA'—OA<A'O. 


REGULAR    POLYGONS   AND    CIRCLES.  217 

But  as  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,  the  length  of  each  side  is  indefinitely  diminished  ; 
that  is,  A'B',  and  consequently  A'C,  is  indefinitely  diminished. 

Therefore  OA'  —  OA,  which  is  less  than  A'C,  is  indefinitely 
diminished. 

Therefore  P—  P  is  indefinitely  diminished.  Q  E  D 

416,  COR.  The  difference  between  the  areas  of  a  regular 
inscribed  polygon  and  of  a  similar  circumscribed  polygon  is 
indefinitely  diminished  as  the  number  of  the  sides  of  the  poly- 
gons is  indefinitely  increased. 

For,  if  /S  and  S1  denote  the  areas  of  the  polygons, 

S'  :  S=  OA'*  :  OA*  =  OA1'  :  00\  §  414 

By  division,  S'  -  S:  S=  OA*2  -  OC*  :  OC\ 

Whence    8'  -8=  8* 


00  OO 

Since  A'C  can  be  indefinitely  diminished  by  increasing  the 
number  of  the  sides,  $'  —  S  can  be  indefinitely  diminished. 

417,  SCHOLIUM.  The  perimeter  P'  is  constantly  greater 
than  P,  and  the  area  S'  is  constantly  greater  than  /S;  for  the 
radius  OA1  is  constantly  greater  than  OA.  But  P'  constantly 
decreases  and  P  constantly  increases  (§  410),  and  the  area  $f 
constantly  decreases,  and  the  area  S  constantly  increases,  as 
the  number  of  sides  of  the  polygons  is  indefinitely  increased. 

Since  the  difference  between  P1  and  P  can  be  made  as 
small  as  we  please,  but  cannot  be  made  absolutely  zero,  and 
since  P1  is  decreasing  while  P  is  increasing,  it  is  evident  that 
P1  and  P  tend  towards  a  common  limit.  This  common  limit 
is  the  length  of  the  circumference.  §  259 

Also,  since  the  difference  between  the  areas  S'  and  S  can  be 
made  as  small  as  we  please,  but  cannot  be  made  absolutely 
zero,  and  since  S9  is  decreasing,  while  S  is  increasing,  it  is 
evident  that  Sf  and  S  tend  towards  a  common  limit.  This 
common  limit  is  the  area  of  the  circle. 


218 


PLANE   GEOMETRY.  —  BOOK   V. 


PROPOSITION  VII.     THEOREM. 

418,  Two  circumferences  have  the  same  ratio  as 
their  radii. 


Let  C  and  C"  be  the  circumferences,  R  and  R'  the 
radii,  of  the  two  circles  Q  and  Q1. 

To  prove  C:  C'=fi:R9. 

Proof,  Inscribe  in  the  CD  two  similar  regular  polygons,  and 
denote  their  perimeters  by  P  and  P. 

Then  P  :  P  -  E  :  E  (§  413)  ;  that  is,  E1  X  P=  E  X  P. 

Conceive  the  number  of  the  sides  of  these  similar  regular 
polygons  to  be  indefinitely  increased,  the  polygons  continuing 
to  have  an  equal  number  of  sides. 

Then  E1  X  P  will  continue  equal  to  Ex  P',  and  P  and  P 
will  approach  indefinitely  O  and  C'  as  their  respective  limits. 

/.  E'xC^ExC1  (§  260)  ;  that  is,  C  :  C'  =  E  :  R'. 

Q.  E.  D. 

419.   COR.   The  ratio  of  the  circumference  of  a  circle  to  its 
diameter  is  constant.    For,  in  the  above  proportion,  by  doubling 
both  terms  of  the  ratio  E  :  E]  ',  we  have 
C:  O1    =  2E:2E'. 

By  alternation,        C'.ZR^C'    :2R'. 

This  constant  ratio  is  denoted  by  IT,  so  that  for  any  circle 
whose  diameter  is  2  E  and  circumference  C,  we  have 


=  ir,   or       = 


420.   SCHOLIUM.    The  ratio  TT  is  incommensurable,  and  there- 
fore can  be  expressed  in  figures  only  approximately. 


REGULAR   POLYGONS   AND   CIRCLES.  219 


PROPOSITION  VIII.     THEOREM. 

421,  The  area  of  a  regular  polygon  is  equal  to  one- 
half  the  product  of  its  apothem  by  its  perimeter. 


A       M 

Let  P  represent  the  perimeter,  R  the  apothem,  and 
8  the  area  of  the  regular  polygon  ABC  etc. 


To  prove  S 

Proof.  Draw  OA,  OB,  OC,  etc. 

The  polygon  is  divided  into  as  many  A  as  it  has  sides. 

The  apothem  is  the  common  altitude  of  these  A, 

and  the  area  of  each  A  is  equal  to  \  R  multiplied  by  the 
base.  §  368 

Hence  the  area  of  all  the  A  is  equal  to  \  R  multiplied  by 
the  sum  of  all  the  bases. 

But  the  sum  of  the  areas  of  all  the  A  is  equal  to  the  area 
of  the  polygon. 

and  the  sum  of  all  the  bases  of  the  A  is  equal  to  the  perim- 
eter of  the  polygon. 

Therefore  S=\RxP. 

Q.  E.  D. 

422.  In  different  circles  similar  arcs,  similar  sectors,  and 
similar  segments  are  such  as  correspond  to  equal  angles  at 
the  centre. 


220  PLANE   GEOMETRY. BOOK   V. 


PROPOSITION  IX.     THEOREM. 

423,  The  area  of  a  circle  is  equal  to  one-half  the 
product  of  its  radius  by  its  circumference. 

E 


B         M        C 

Let  E   represent   the  radius,    C  the  circumference, 
and  &  the  area,  of  the  circle. 


To  prove  S 

Proof,  Circumscribe  any  regular  polygon  about  the  circle, 
and  denote  its  perimeter  by  P. 

Then  the  area  of  this  polygon  =  |  E  X  P,       §  421 

Conceive  the  number  of  sides  of  the  polygon  to  be  indefi- 
nitely increased  ;  then  the  perimeter  of  the  polygon  approaches 
the  circumference  of  the  circle  as  its  limit,  and  the  area  of  the 
polygon  approaches  the  circle  as  its  limit. 

But  the  area  of  the  polygon  continues  to  be  equal  to  one- 
half  the  product  of  the  radius  by  the  perimeter,  however  great 
the  number  of  sides  of  the  polygon.  / 

Therefore  S=%£  X(J.  §  260 

Q.  E.  D. 

424,  COR.  1.    The  area  of  a  sector  equals  one-half  the  product 
of  its  radius  by  its  arc.     For  the  sector  is  such  a  part  of  the 
circle  as  its  arc  is  of  the  circumference. 

425,  COR.  2.    The  area  of  a  circle  equals  TT  times  the  square 
of  its  radius. 

For  the  area  of  the  O  = 


REGULAR   POLYGONS   AND   CIRCLES.  221 

426.  COE.  3.  The  areas  of  two  circles  are  to  each  other  as  the 
squares  of  their  radii.  For,  if  8  and  /S"  denote  the  areas,  and 
E  and  JR'  the  radii, 


427,  COE.  4.  /Similar  arcs,  being  like  parts  of  their  respective 
circumferences,  are  to  each  other  as  their  radii  ;  similar  sectors, 
being  like  parts  of  their  respective  circles,  are  to  each  other  as 
the  squares  of  their  radii. 

PEOPOSITION  X.     THEOEEM. 

428.  The  areas  of  two  similar  segments  are  to  each 
other  as  the  squares  of  their  radii. 


P' 

Let  AC  andA'C1  be  the  radii  of  the  two  similar  seg- 
ments ASP  and  A'B'P. 

To  prove          ABP  :  A'B'P  =  AC*  :  A^. 

Proof.     The  sectors  ACE  and  A'C'E'  are  similar,         §  422 
(having  the  A  at  the  centre,  C  and  C',  equal). 


/.C=/.  C',  A0=  CB,  and  A'C'  =  C'£'. 

Therefore  the  A  ACE  and  A'C'&  are  similar.  §  326 

Now     sector  ACE  :  sector  A'C'£'  =  AC2  :  A'C'\  §  427 

and  AAC£:AA'C'£'  =  AG2:I'C''\  §375 

Hence  sector  ACB~  A  ACB     =*&.  §301 

sector  A'C'£'-A  A'C'3'      J_>c<* 

That  is,          AEP  :  A'  &P*  =  AC*  :  J7(^2. 

Q.E.O. 


222 


PLANE   GEOMETRY.  —  BOOK   V. 


PROBLEMS  OF  CONSTRUCTION. 
PROPOSITION  XI.     PROBLEM. 
429.  To  inscribe  a  square  in  a  given  circle. 


A 


Let  0  be  the  centre  of  the  given  circle. 

To  inscribe  a  square  in  the  circle. 

Construction,  Draw  the  two  diameters  AC  and  BD  _L  to 
each  other. 

Join  AB,  BC,  CD,  and  DA. 

Then  ABCD  is  the  square  required. 

Proof,          The  A  ABC,  BCD,  etc.,  are  rt,  A,  §  264 

(being  inscribed  in  a  semicircle), 

and  the  sides  AB,  BC,  etc.,  are  equal,  §  230 

(in  the  same  O  equal  arcs  are  subtended  by  equal  chords). 

Hence  the  figure  ABCD  is  a  square.  §  171 

Q.  E.  F. 

430,  COR.  By  bisecting  the  arcs  AB,  BC,  etc.,  a  regular 
polygon  of  eight  sides  may  be  inscribed  in  the  circle ;  and,  by 
continuing  the  process,  regular  polygons  of  sixteen,  thirty-two, 
sixty-four,  etc.,  sides  may  be  inscribed. 


Ex.  376.  The  area  of  a  circumscribed  square  is  equal  to  twice  the 
area  of  the  inscribed  square. 

Ex.  377.  If  the  length  of  the  side  of  an  inscribed  square  is  2  inches, 
what  is  the  length  of  the  circumscribed  square  ? 


PROBLEMS   OF   CONSTRUCTION.  223 

PROPOSITION  XII.     PROBLEM. 
431,  To  inscribe  a  regular  hexagon  in  a  given  circle. 


Let  0  be  the  centre  of  the  given  circle. 

To  inscribe  in  the  given  circle  a  regular  hexagon. 
Construction,    From  0  draw  any  radius,  as  OO. 

From  C"as  a  centre,  with  a  radius  equal  to  00, 
describe  an  arc  intersecting  the  circumference  at  F. 

Draw  0_Fand  OF. 

Then  OF  is  a  side  of  the  regular  hexagon  required. 
Proof,    The  A  OFC  is  equilateral  and  equiangular. 
Hence  the  Z  FOC  is  £  of  2  rt.  A,  or  £  of  4  rt.  A          §  138 
And  the  arc  FCis  ^  of  the  circumference  ABCF. 
Therefore  the  chord  FC,  which  subtends  the  arc  FCt  is  a 
side  of  a  regular  hexagon  ; 

and  the  figure  OFD  etc.,  formed  by  applying  the  radius  six 
times  as  a  chord,  is  a  regular  hexagon.  Q.E.  F. 

432,  COR.  1.  By  joining  the  alternate  vertices  A,  O,  D,  an 
equilateral  triangle  is  inscribed  in  the  circle. 

433,  COR.  2.  By  bisecting  the  arcs  AB,  BO,  etc.,  a  regular 
polygon  of  twelve  sides  may  be  inscribed  in  the  circle ;  and,  by 
continuing  the  process,  regular  polygons  of  twenty-four,  forty- 
eight,  etc.,  sides  may  be  inscribed, 


224 


PLANE   GEOMETRY.  —  BOOK    V. 


PROPOSITION  XIII.     PROBLEM. 
434,  To  inscribe  a  regular  decagon  in  a  given  circle. 


Let  0  be  the  centre  of  the  given  circle. 

To  inscribe  a  regular  decagon  in  the  given  circle. 

Construction,  Draw  the  radius  0(7, 

and  divide  it  in  extreme  and  mean  ratio,  so  that  OO  shall 
be  to  0/S.as  OS  is  to  SO.  §  356 

From  O  as  a  centre,  with  a  radius  equal  to  OS, 

describe  an  arc  intersecting  the  circumference  at  B,  and 
draw  BO. 

Then  EG  is  a  side  of  the  regular  decagon  required. 

Proof,  Draw  BS  and  BO. 

By  construction        OC  :  OS  =  OS  :  SO, 
and  BC=OS. 


Moreover,  the  Z  OCB  =  Z.  SOB.  Iden. 

Hence  the  A  OCB  and  BCS  we  similar,         §  326 
(having  an  Z.  of  the  one  equal  to  an  Z.  of  the  other,  and  the  including  sides 
proportional). 

But  the  A  OCB  is  isosceles, 

(its  sides  00  and  OB  being  radii  of  the  same  circle). 

.'.  A  BOS,  which  is  similar  to  the  A  OOB,  is  isosceles, 

and  CB^BS^OS. 


PROBLEMS   OF   CONSTRUCTION.  225 

/.  the  A  SOB  is  isosceles,  and  the  Z  0  =-  Z  SBO. 
But  the  ext.  Z  CSB  =  Z  0  +  Z  8BO  =  2  Z  0.    §145 
Hence  Z  SOB  (=  Z  C&5)  -  2  Z  0,  §  154 

and  Z  0£<7(=  Z  #<?£)  =  2  Z  0.  §  154 

.'.  the  sum  of  the  A  of  the  A  OCB  =  5  Z  0  =  2  rt.  z§, 
and  Z  0  —  |  of  2  rt.  A,  or  ^  of  4  rt.  A. 

Therefore  the  arc  BCia  -£$  of  the  circumference, 
and  the  chord  BO  is  a  side  of  a  regular  inscribed  decagon. 

Hence,  to  inscribe  a  regular  decagon,  divide  the  radius  in 
extreme  and  mean  ratio,  and  apply  the  greater  segment  ten 
times  as  a  chord. 

Q.E.F. 

435.  COR.  1.  By  joining  the  alternate  vertices  of  a  regular 
inscribed  decagon,  a  regular  pentagon  is  inscribed. 

436,  COR.  2.  By  bisecting  the  arcs  BC,  OF,  etc.,  a  regular 
polygon  of  twenty  sides  may  be  inscribed ;  and,  by  continuing 
the  process,  regular  polygons  of  forty,  eighty,  etc.,  sides  may  be 
inscribed. 


Let  R  denote  the  radius  of  a  regular  inscribed  polygon,  r  the  apothem, 
a  one  side,  A  an  interior  angle,  and  0  the  angle  at  the  centre  ;  show  that 

Ex.  378.   In  a  regular  inscribed  triangle  a  =  JB  V3,  r  =  J  £,  A=*  60°, 
C=  120°. 


Ex.379.   In  an   inscribed  square   a  =  ftV2,   r  =  %R^2,   A  =  90°, 
=90°. 

Ex.  380.   In  a  regular  inscribed  hexagon  a  =  R,  r  =  J  R  V3,  A  =  120°, 
=  60°. 

Ex.  381.   In  a  regular  inscribed  decagon 

,    4=144°,    (7=36°. 


226  PLANE   GEOMETEY.  —  BOOK   V. 


PROPOSITION  XIV.     PROBLEM. 

437,  To  inscribe  in  a  given  circle  a  regular  pente- 
decagon,  or  polygon  of  fifteen  sides. 


F 
Let  Q  be  the  given  circle. 

To  inscribe  in  Q  a  regular  pentedecagon. 

Construction,  Draw  EH  equal  to  a  side  of  a  regular  inscribed 
hexagon,  §  431 

and  EF equal  to  a  side  of  a  regular  inscribed  decagon.  §  434 

Join  FH. 

Then  _F_ZTwill  be  a  side  of  a  regular  inscribed  pentedecagon. 
Proof,        The  arc  EH\$  \  of  the  circumference-, 

and  the  arc  EFis  y1^  of  the  circumference. 
Hence  the  arc  FH  \&  -J-  —  y1^-,  or  y1^,  of  the  circumference, 

and  the  chord  FH  is  a  side  of  a  regular  inscribed  pente- 
decagon. 

By  applying   FH  fifteen  times  as  a  chord,  we  have  the 
polygon  required. 

Q.E.  F. 

438,  COR.  By  bisecting  the  arcs  FH,  HA,  etc.,  a  regular 
polygon  of  thirty  sides  may  be  inscribed ;  and,  by  continuing 
the  process,  regular  polygons  of  sixty,  one  hundred  twenty,  etc., 
sides,  may  be  inscribed. 


PROBLEMS   OF   CONSTRUCTION.  227 


PROPOSITION  XV.     PROBLEM. 

439,  To  inscribe  in  a  given  circle  a  regular  polygon 
similar  to  a  given  regular  polygon. 


Let  ABCD  etc.,  be  the  given  regular  polygon,  and 
OD'E'  the  given  circle. 

To  inscribe  in  the  circle  a  regular  polygon  similar  to  A£CD, 
etc. 

Construction,    From  0,  the  centre  of  the  given  polygon, 

draw  OD  and  00. 
From  0',  the  centre  of  the  given  circle, 

draw  O'C'  and  O'D', 
making  the  Z  0' =  Z  0. 

Draw  C'D'. 
Then  C'D1  will  be  a  side  of  the  regular  polygojyjequired. 

Proof,    Each  polygon  will  have  as  Eianyi|H       Kthe  Z  0 
(=  Z  0')  is  contained  times  in  4  rt.  A. 

Therefore  the  polygon  C'D'jE'  etc.,  is  similar  to  the  poly- 
gon CDE  etc.,  "**"*'§  411 
(two  regular  polygons  of  the  same  number  of  sides  are  similar). 

Q.  E.  F. 


228  PLANE   GEOMETRY.  —  BOOK   V. 


PROPOSITION  XVI.     PROBLEM. 

440,  Given  the  radius  and  the  side  of  a  regular 
inscribed  polygon,  to  find  the  side  of  the  regular 
inscribed  polygon  of  double  the  number  of  sides. 

D 


H 
LetAB  be  a  side  of  the  regular  inscribed  polygon. 

To  find  the  value  of  AD,  a  side  of  a  regular  inscribed  poly- 
gon of  double  the  number  of  sides. 

From  D  draw  DH  through  the  centre  0,  and  draw  OA,  AH. 

DHis  J_  to  AB  at  its  middle  point  0.  §  123 

In  the  rt.  A  0,4  £,  OC*=  OA*  -  AC\  §339 


That  is,  Otf=V<5Z*~  AC\ 

But  AQ=\AB\  hence 


Therefore,  OO=  ^  OA*  -\AB\ 

In  the  rt.  A  DAtf,  §  264 

§334 


and  AD  -  -\i20A(OA  -  00). 


If  we  denote  the  radius  by  E,  and  substitute  Vj?2  — 
for  0(7,  then 


AD  = 


=  ^JR  (2  E-  V4  B*  -  AB\ 


Q.E.F. 


PROBLEMS   OF   COMPUTATION. 


229 


PROPOSITION  XVII.    PROBLEM. 

441,  To  compute  the  ratio  of  the  circumference  of  a 
circle  to  its  diameter  approximately. 


Let  C  be  the  circumference,  and  R  the  radius. 

To  find  the  numerical  value  of  IT. 


0.  §419 

Therefore  when  R  —  1,  IT  =  %  0. 

We  make  the  following  computations  by  the  use  of  the 
formula  obtained  in  the  last  proposition,  when  .22=1,  and 
AB=\  (a  side  of  a  regular  hexagon). 


No. 
Bides. 


Form  of  Computation. 


Length  of  Side.       Length  of  Perimeter. 


12  Cl=' 
24  c.=- 


0.51763809, 
0.26105238 


6.21165708 
6,26525722 


-V4~-  (0.517638097 

48  c,  -  V2  -  Vf-  (0.261052387  0.13080626  6.27870041 

96  <?4=  V^_  V4-  (0.130806H7  0.06543817  6.28206396 

192  c,  =  V2  -  VfHT)706543817y  0.03272346  6.28290510 

0.01636228  6.28311544 

0.00818121  6.28316941 


384  c6=  V2-  V4-(0.032723H7 
768  c7=V2-V4~_  (0.016362287 


Hence  we  may  consider  6.28317  as  approximately  the  cir- 
cumference of  a  O  whose  radius  is  unity. 

Therefore  IT  =  £(6.28317)  =  3.14159  nearly.       Q.E.F. 
442,   SCHOLIUM.   In  practice,  we  generally  take 
a-  =  3.1416,     1-0.31831. 

7T 


230 


PLANE   GEOMETRY. BOOK   V. 


MAXIMA  AND  MINIMA. — SUPPLEMENTARY. 

443,  Among  magnitudes  of  the  same  kind,  that  which  is 
greatest  is  the  maximum,  and  that  which  is  smallest  is  the 
minimum. 

Thus  the  diameter  of  a  circle  is  the  maximum  among  all 
inscribed  straight  lines  ;  and  a  perpendicular  is  the  minimum 
among  all  straight  lines  drawn  from  a  point  to  a  given  line. 

444,  Isoperimetric  figures  are    figures  which   have  equal 
perimeters. 

PROPOSITION  XVIII.     THEOREM. 

445,  Of  all  triangles  having  two  given  sides,  that 
in  which  these  sides  include  a  right  angle  is  the 

maximum. 

A. 
E  K        E 


Let  the  triangles  ABC  and  EEC  have  the  sides  AB 
and  BC  equal  respectively  to  EB  and  BC ;  and  let  the 
angle  ABC  be  a  right  angle. 

To  prove  A  ABO > A  EEC. 

Proof,  From  E  let  fall  the  ±  ED. 

The  A  ABC  and  EEC,  having  the  same  base  BC,  are  to 
each  other  as  their  altitudes  AB  and  ED.  §370 

Now  EB  >  ED.  §  114 

By  hypothesis,  EB  =  AB. 

/.  AB  >  ED. 


MAXIMA   AND    MINIMA. 


231 


PROPOSITION  XTX.     THEOREM. 

446,  Of  all  triangles  having  the  same  base  and  equal 
perimeters,  the  isosceles  triangle  is  the  maximum. 


Let  the  A  ACS  and  ADB  have  equal  perimeters,  and 
let  the  A  ACB  be  isosceles. 

To  prove  AACB>A  ADB. 

Proof,    Produce  AC  to  H,  making  CH=  AC,  and  join  HE. 
ABH\&  a  right  angle,  for  it  will  be  inscribed  in  the  semi- 
circle whose  centre  is  (7,  and  radius  CA. 

Produce  HE,  and  take  DP  =  DB. 

Draw  CKsiud  DM  II  to  AS,  and  join  AP. 

Now  AH=  AC+  CB=--AD  +  DB  =  AD  +  DP. 

EutAD  +  DP>AP,  hence  AH>  AP. 

Therefore  HE  >  BP.  §  120 

But  KE^IHE  and  ME=\EP.  §121 

Hence  KB>  MB. 

By  §  180,  KB  =  CE  and  MB=DF,  the  altitudes  of  the 
&ACB  and  ADB. 

Therefore  A  ABC>  A  ADB.  §  370 

Q.  E.  D. 


232  PLANE  GEOMETRY.  —  BOOK  V. 


PROPOSITION  XX.     THEOREM. 

447,  Of  all  polygons  with  sides  all  jj>iven  but  one, 
the  maximum  can  be  inscribed  in  a  semicircle  which 
has  the  undetermined  side  for  its  diameter. 

c 


Let  ABODE  be  the  maximum  of  polygons  with  sides 
AB,  BCt  CD,  DE,  and  the  extremities  A  and  E  on  the 
straight  line  MN. 

To  prove  ABCDE  can  be  inscribed  in  a  semicircle. 

Proof,    From  any  vertex,  as  O}  draw  CA  and  CE. 

The  A  ACE  must  be  the  maximum  of  all  A  having  the 
given  sides  CA  and  CE\  otherwise,  by  increasing  or  diminish- 
ing the  Z  ACE,  keeping  the  sides  CA  and  CE  unchanged,  but 
sliding  the  extremities  A  and  E  along  the  line  MN,  we  can 
increase  the  A  AGE,  while  the  rest  of  the  polygon  will  remain 
'  unchanged,  and  therefore  increase  the  polygon. 

But  this  is  contrary  to  the  hypothesis  that  the  polygon  is 
the  maximum  polygon. 

Hence  the  A  ACE  with  the  given  sides  CA  and  CE  is  the 

maximum. 

Therefore  the  Z  ACE  is  a  right  angle,  §  445 

(the  maximum  of  A  having  two  given  sides  is  the  A  with  the  two  given  sides 
including  a  rt.  Z). 

Therefore  C  lies  on  the  semi-circumference.        §  264 

Hence  every  vertex  lies  on  the  circumference ;  that  is,  the 

maximum  polygon  can  be  inscribed  in  a  semicircle  having  the 

undetermined  side  for  a  diameter.      ,  o.  E.  D. 


MAXIMA   AND   MINIMA. 


233 


PROPOSITION  XXI.     THEOREM. 

448,   Of  all  polygons  ivith  given  sides,  that  which 
can  ~be  inscribed  in  a  circle  is  the  maximum. 


H'     D' 


Let  ABODE  be  a  polygon  inscribed  in  a  circle,  and 
A'B'C'D'E'  be  a  polygon,  equilateral  with  respect  to 
ABODE,  which  cannot  be  inscribed  in  a  circle. 

To  prove     ABCDE  greater  than  A'3'C'D'E'. 

Proof.  Draw  the  diameter  AIT. 

Join  Off  and  DH. 
Upon  C'D'  (=  CD)  construct  the  A  C'H'D'  =  A  CHD, 

and  draw  A1  If1. 

Now  ABCH>  A'3'C'JI1,  §  447 

and  AEDH  >  A'E'D'H', 

(of  all  polygons  with  sides  all  given  but  one,  the  maximum  can  be  inscribed 
in  a  semicircle  having  the  undetermined  side  for  its  diameter). 

Add  these  two  inequalities,  then 
ABCHDE  >  A'&C'H'D'U'. 

Take  away  from  the  two  figures  the  equal  A  CUD  and  C* 
Then  ABODE  >  A'3'C'D'E'. 


234 


PLANE   GEOMETRY.  —  BOOK   V. 


PROPOSITION  XXII.     THEOREM. 

449,  Of  isoperimetric  polygons  of  the  same  number 
of  sidest  the  maximum  is  equilateral. 

K 


Let  ABCD   etc.,   be   the  maximum   of  isoperimetric 
polygons  of  any  given  number  of  sides. 

To  prove  AB,  EC,  CD,  etc.,  equal. 

Proof,  Draw  AC. 

The  A  ABC  must  be  the  maximum  of  all  the  A  which  are 
formed  upon  AC  with  a  perimeter  equal  to  that  of  A  ABC. 

Otherwise,    a   greater   A  AKC  could   be   substituted   for 
A  ABO,  without  changing  the  perimeter  of  the  polygon. 

But  this  is  inconsistent  with  the  hypothesis  that  the  poly- 
gon ABCD  etc.,  is  the  maximum  polygon. 

.-.  the  A  ABC  is  isosceles,  §  446 

(of  all  A  having  the  same  base  and  equal  perimeters,  the  isosceles  A  is  the 
maximum). 

In  like  manner  it  may  be  proved  that  BC=  CD,  etc.  Q.E.D. 
450,    COR.    The  maximum  of  isoperimetric  polygons  of  the 
same  number  of  sides  is  a  regular  polygon. 

For,  it  is  equilateral,  §  449 

(the  maximum  of  isoperimetric  polygons  of  the  same  number  of  sides  is 
equilateral). 

Also  it  can  be  inscribed  in  a  circle,  §  448 

(the  maximum  of  all  polygons  formed  of  given  sides  can  be  inscribed  in  a  O). 
That  is,  it  is  equilateral  and  equiangular, 

and  therefore  regular.         .  §  395 


MAXIMA  AND   MINIMA. 


235 


PROPOSITION  XXIII.    THEOREM. 

451,  Of  isoperimetric  regular  polygons,  that  which 
has  the  greatest  number  of  sides  is  the  maximum, 
C 


A      ^D        B 

Let  Q  be  a  regular  polygon  of  three  sides,  and  $ 
a  regular  polygon  of  four  sides,  and  let  the  two  poly- 
gons have  equal  perimeters. 

To  prove  Qr  greater  than  Q. 

Proof,       Draw  CD  from  C  to  any  point  in  AB. 

Invert  the  A  CD  A  and  place  it  in  the  position  DCE,  let- 
ting D  fall  at  (7,  C  at  D,  and  A  at  E. 

The  polygon  DECE  is  an  irregular  polygon  of  four  sides, 
which  by  construction  has  the  same  perimeter  as  Q1,  and  the 
same  area  as  Q. 

Then  the  irregular  polygon  DECE  of  four  sides  is  less  than 
the  regular  isoperimetric  polygon  Q'  of  four  sides.  §  450 

In  like  manner  it  may  be  shown  that  Qf  is  less  than  a  regular 
isoperimetric  polygon  of  five  sides,  and  so  on.  Q.  E.  D. 

452,  COR.  The  area  of  a  circle  is  greater  than  the  area  of 
any  polygon  of  equal  perimeter. 

382.  Of  all  equivalent  parallelograms  having  equal  bases,  the  rec- 
tangle has  the  least  perimeter. 

383.  Of  all  rectangles   of  a  given  area,  the  square  has  the  least 
perimeter. 

384.  Of  all  triangles  upon  the  same  base,  and  having  the  same  alti- 
tude, the  isosceles  has  the  least  perimeter. 

v    385.  To  divide  a  straight  line  into  two  parts  such  that  their  product 
shall  be  a  maximum. 


236 


PLANE   GEOMETRY.  —  BOOK   V. 


PROPOSITION  XXIV.     THEOREM. 

453,  Of  regular  polygons  having  a  given  area,  that 
which  has  the  greatest  number  of  sides  has  the  least 
perimeter. 


Let  Q  and  Q1  be  regular  polygons  having  the  same 
area,  and  let  Q'  have  the  greater  number  of  sides. 

To  prove  the  perimeter  of  Q,  greater  than  the  perimeter  of  Qf. 

Proof,    Let  Qn  be  a  regular  polygon  having  the  same  perim- 
eter as  Q',  and  the  same  number  of  sides  as  Q. 

Then  Q'  >  Q",  §  451 

(of  isoperimetric  regular  polygons,  that  which  has  the  greatest  number  of 
sides  is  the  maximum). 


But 


.-.  Q  >  Q"> 

/.  the  perimeter  of  Q  >  the  perimeter  of  Qn. 
But  the  perimeter  of  Q'  —  the  perimeter  of  Q". 
.'.  the  perimeter  of  Q  >  that  of  Q1. 


Cons. 


Q.E.  D. 


454.    Con.    The  circumference  of  a  circle   is   less   than  the 
perimeter  of  any  polygon  of  equal  area. 


386.  To  inscribe  in  a  semicircle  a  rectangle  having  a  given  area; 
a  rectangle  having  the  maximum  area. 

387.  To  find  a  point  in  a  semi-circumference  such  that  the  sum  of  its 
distances  from  the  extremities  of  the  diameter  shall  be  a  maximum. 


EXERCISES.  237 


THEOREMS. 

388.  The  side  of  a  circumscribed  equilateral  triangle  is  equal  to  twice 
the  side  of  the  similar  inscribed  triangle.    Find  the  ratio  of  their  areas. 

389.  The  apothem  of  an  inscribed  equilateral  triangle  is  equal  to  half 
the  radius  of  the  circle. 

390.  The  apothem  of  an  inscribed  regular  hexagon  is  equal  to  half 
the  side  of  the  inscribed  equilateral  triangle. 

391.  The  area  of  an  inscribed  regular  hexagon  is  equal  to  three- 
fourths  that  of  the  circumscribed  regular  hexagon. 

392.  The  area  of  an  inscribed  regular  hexagon  is  a  mean  proportional 
between  the  areas  of  the  inscribed  and  the  circumscribed  equilateral 
triangles. 

393.  The  area  of  an  inscribed  regular  octagon  is  equal  to  that  of  a 
rectangle  whose  sides  are  equal  to  the  sides  of  the  inscribed  and  the  cir- 
cumscribed squares. 

394.  The  area  of  an  inscribed  regular  dodecagon  is  equal  to  three 
times  the  square  of  the  radius. 

_j  395.   Every  equilateral  polygon  circumscribed  about  a  circle  is  regu- 
lar if  it  has  an  odd  number  of  sides. 

396.  Every  equiangular  polygon  inscribed  in  a  circle  is  regular  if  it 
has  an  odd  number  of  sides. 

397.  Every  equiangular  polygon   circumscribed  about  a  circle   is 
regular. 

398.  Upon  the  six  sides  of  a  regular  hexagon  squares  are  constructed 
outwardly.   Prove  that  the  exterior  vertices  of  these  squares  are  the  ver- 
tices of  a  regular  dodecagon. 

399.  The  alternate  vertices  of  a  regular  hexagon  are  joined  by  straight 
lines.    Prove  that  another  regular  hexagon  is  thereby  formed.    Find  the 
ratio  of  the  areas  of  the  two  hexagons. 

400.  The  radius  of  an  inscribed  regular  polygon  is  the  mean  propor- 
tional between4  its  apothem  and  the  radius  of  the  similar  circumscribed 
regular  polygon. 

401.  The  area  of  a  circular  ring  is  equal  to  that  of  a  circle  whose 
diameter  is  a  chord  of  the  outer  circle  and  a  tangent  to  the  inner  circle. 

402.  The  square  of  the  side  of  an  inscribed  regular  pentagon  is  equal 
to  the  sum  of  the  squares  of  the  radius  of  the  circle  and  the  side  of  the 
inscribed  regular  decagon. 


238  PLANE   GEOMETRY.  —  BOOK    V. 

If  R  denotes  the  radius  of  a  circle,  and  a  one  side  of  a  regular  inscribed 
polygon,  show  that : 

r>        

403.   In  a  regular  pentagon,  a  =  —  rv/10  — 2V5. 


404.  In  a  regular  octagon,     a  =  fi  -\/2  —  V2. 

405.  In  a  regular  dodecagon,  a  =  R  V2  —  V3. 

406.  If  on  the  legs  of  a  right  triangle,  as  diameters,  semicircles  are 
described  external  to  the  triangle,  and  from  the  whole  figure  a  semicircle 
on  the  hypotenuse  is  subtracted,  the  remainder  is  equivalent  to  the  given 
triangle. 

NUMERICAL  EXERCISES. 

407.  The  radius  of  a  circle  =r.     Find  one  side  of  the  circumscribed 
equilateral  triangle.        ^  A/} 

408.  The  radius  of  a  circle'  =  r.     Find  one  side  of  the  circumscribed 
regular  hexagon.     ukrVJ 

409.  If  the  radius  of  a  circle  is  r,  and  the  side  of  an  inscribed  regular 
polygon  is  a,  show  that  the  side  of  the  similar  circumscribed  regular 
polygon  is  equal  to  2ar          ^ 

V4r2-a2 

410.  The  radius  of  a  circle  =  r.     Prove  that  the  area  of  the  inscribed 
regular  octagon  is  equal  to  2r2v/2.jL- 

411.  The  sides  of  three  regular  octagons  are  3  feet,  4  feet,  and  5  feet, 
respectively.     Find  the  side  of  a  regular  octagon  equal  in  area  to  the 
sum  of  the  areas  of  the  three  given  octagons. 

412.  What  is  the  width  of  the  ring  between  two  concentric  circum- 
ferences whose  lengths  are  440  feet  and  330  feet? 

'  413.   Find  the  angle  subtended  at  the  centre  by  an  arc  6  feet  5  inches 
long,  if  the  radius  of  the  circle  is  8  feet  2  inches. 

414.  Find  the  angle  subtended  at  the  centre  of  a  circle  by  an  arc 
whose  length  is  equal  to  the  radius  of  the  circle. 

415.  What  is  the  length  of  the  arc  subtended  by  one  side  of  a  regular 
dodecagon  inscribed  in  a  circle  whose  radius  is  14  feet  ? 

416.  Find  the  side  of  a  square  equivalent  to  a  circle  whose  radius  is 
56  feet. 


EXERCISES.  239 

417.  Find  the  area  of  a  circle  inscribed  in  a  square  containing  196 
square  feet. 

418.  The  diameter  of  a  circular  grass  plot  is  28  feet.     Find  the  diam- 
eter of  a  circular  plot  just  twice  as  large. 

419.  Find  the  side  of  the  largest  square  that  can  be  cut  out  of  a  cir- 
cular piece  of  wood  whose  radius  is  1  foot  8  inches. 

420.  The  radius  of  a  circle  is  3  feet.    What  is  the  radius  of  a  circle  25 
times  as  large  ?   J  as  large  ?   ^  as  large  ? 

421.  The  radius  of  a  circle  is  9  feet.     "What  are  the  radii  of  the  con- 
centric circumferences  that  will  divide  the  circle  into  three  equivalent 
parts? 

22.   The  chord  of  half  an  arc  is  12  feet,  and  the  radius  of  the  circle  is 
i;t.     Find  the  height  of  the  arc. 

423.  The  chord  of  an  arc  is  24  inches,  and  the  height  of  the  arc  is  9 
inches.     Find  the  diameter  of  the  circle. 

424.  Find  the  area  of  a  sector,  if  the  radius  of  the  circle  is  28  feet, 
and  the  angle  at  the  centre  22^°. 

425.  The  radius  of  a  circle  =  r.     Find  the  area  of  the  segment  sub- 
tended by  one  side  of  the  inscribed  regular  hexagon. 

426.  Three  equal  circles  are  described,  each  touching  the  other  two. 
If  the  common  radius  is  r,  find  the  area  contained  between  the  circles. 

PROBLEMS. 

To  circumscribe  about  a  given  circle  : 

427.  An  equilateral  triangle.  429.   A  regular  hexagon. 
428..  A  square.                                    430.   A  regular  octagon. 

431.   To  draw  through  a  given  point  a  line  so  that  it  shall  divide  a 
given  circumference  into  two  parts  having  the  ratio  3 :  7. 

132.   To  construct  a  circumference  equal  to  the  sum  of  two  given 
•iimferences. 

433.  To  construct  a  circle  equivalent  to  the  sum  of  two  given  circles. 

434.  To  construct  a  circle  equivalent  to  three  times  a  given  circle. 

435.  To  construct  a  circle  equivalent  to  three-fourths  of  a  given  circle. 
To  divide  a  given  circle  by  a  concentric  circumference : 

436.  Into  two  equivalent  parts.        437.   Into  five  equivalent  parts. 


240  PLANE   GEOMETRY. BOOK   V. 

MISCELLANEOUS  EXERCISES. 

THEOREMS. 

438.  The  line  joining  the  feet  of  the  perpendiculars  dropped  from  the 
extremities  of  the  base  of  an  isosceles  triangle  to  the  opposite  sides  is 
parallel  to  the  base. 

439.  If  AD  bisect  the  angle  A  of  &  triangle  ABC,  and  BD  bisect  the 
exterior  angle  CBF,  then  angle  ADB  equals  one-half  angle  AGE. 

440.  The  sum  of  the  acute  angles  at  the  vertices  of  a  pentagram  (five- 
pointed  star)  is  equal  to  "two  right  angles. 

441.  The  bisectors  of  the  angles  of  a  parallelogram  form  a  rectangle. 

442.  The  altitudes  AD,  BE,  CF  of  the  triangle  ABC  bisect  the  angles 
of  the  triangle  DEF. 

HINT.   Circles  with  AB,  BC,  AC  as  diameters  will  pass  through  E  and 
D,  E  and  F,  D  and  Ft  respectively. 

443.  The  portions  of  any  straight  line  intercepted  between  the  cir- 
cumferences of  two  concentric  circles  are  equal. 

444.  Two  circles  are  tangent  internally  at  P,  and  a  chord  AB  of  the 
larger  circle  touches  the  smaller  circle  at  C.     Prove  that  PC  bisects  the 
angle  APB. 

HINT.   Draw  a  common  tangent  at  P,  and  apply  §$  263,  269,  145. 

445.  The  diagonals  of  a  trapezoid  divide  each  other  into  segments 
which  are  proportional. 

446.  The  perpendiculars  from  two  vertices  of  a  triangle  upon  the 
opposite  sides  divide  each  other  into  segments  reciprocally  proportional. 

447.  If  through  a  point  P  in  the  circumference  of  a  circle  two  chords 
are  drawn,  the  chords  and  the  segments  between  P  and  a  chord  parallel 
to  the  tangent  at  Pare  reciprocally  proportional. 

448.  The  perpendicular  from  any  point  of  a  circumference  upon  a 
chord  is  a  mean  proportional  between  the  perpendiculars  from  the  same 
point  upon  the  tangents  drawn  at  the  extremities  of  the  chord. 

449.  In  an  isosceles  right  triangle  either  leg  is  a  mean  proportional 
between  the  hypotenuse  and  the  perpendicular  upon  it  from  the  vertex 
of  the  right  angle. 

450.  The  area  of  a  triangle  is  equal  to  half  the  product  of  its  perim- 
eter by  the  radius  of  the  inscribed  circle. 


MISCELLANEOUS   EXERCISES.  241 

451.  The  perimeter  of  a  triangle  is  to  one  side  as  the  perpendicular 
from  the  opposite  vertex  is  to  the  radius  of  the  inscribed  circle. 

452.  The  sum  of  the  perpendiculars  from  any  point  within  a  convex 
equilateral  polygon  upon  the  sides  is  constant. 

453.  A  diameter  of  a  circle  is  divided  into  any  two  parts,  and  upon 
these  parts  as  diameters  semi-circumferences  are  described  on  opposite 
sides  of  the  given  diameter.    Prove  that  the  sum  of  their  lengths  is  equal 
to  the  semi-circumference  of  the  given  circle,  and  that  they  divide  the 
circle  into  two  parts  whose  areas  have  the  same  ratio  as  the  two  parts 
into  which  the  diameter  is  divided. 

454.  Lines  drawn  from  one  vertex  of  a  parallelogram  to  the  middle 
points  of  the  opposite  sides  trisect  one  of  the  diagonals, 

455.  If  two  circles  intersect  in  the  points  A  and  J5,  and  through  A 
any  secant  CAD  is  drawn  limited  by  the  circumferences  at  0  and  D,  the 
straight  lines  EG,  ED,  are  to  each  other  as  the  diameters  of  the  circles. 

456.  If  three  straight  lines  AA/  ',  BE',  CO',  drawn  from  the  vertices 
of  a  triangle  ABC  to  the  opposite  sides,  pass  through  a  common  point  0 
within  the  triangle,  then 

OA'      OB'     PC'  =  l 
AA'     B&      CCf 


457.  Two  diagonals  of  a  regular  pentagon,  not  drawn  from  a  common 
vertex,  divide  each  other  in  extreme  and  mean  ratio. 

I 

Loci. 

458.  Find  the  locus  of  a  point  P  whose  distances  from  two  given 
points  A  and  B  are  in  a  given  ratio  (m  \  n). 

459.  OP  is  any  straight  line  drawn  from  a  fixed  point  O  to  the  cir- 
cumference of  a  fixed  circle  ;  in  OP  a  point  Q  is  taken  such  that  OQ:  OP 
is  constant.    Find  the  locus  of  Q. 

460.  From  a  fixed  point  A  a  straight  line  AB  is  drawn  to  any  point 
in  a  given  straight  line  CD,  and  then  divided  at  P  in  a  given  ratio 
(m  :  n).    Find  the  locus  of  the  point  P. 

461.  Find  the  locus  of  a  point  whose  distances  from  two  given  straight 
lines  are  in  a  given  ratio.     (The  locus  consists  of  two  straight  lines.) 

462.  Find  the  locus  of  a  point  the  sum  of  whose  distances  from  two 
given  straight  lines  is  equal  to  a  given  length  L    (See  Ex,  73.) 


242  PLANE   GEOMETRY.  —  BOOK   Y. 


PROBLEMS. 

463.  Given  the  perimeters  of  a  regular  inscribed  and  a  similar  circum- 
scribed polygon,  to  compute  the  perimeters  of  the  regular  inscribed  and 
circumscribed  polygons  of  double  the  number  of  sides. 

464.  To  draw  a  tangent  to  a  given  circle  such  that  the  segment  inter- 
cepted between  the  point  of  contact  and  a  given  straight  line  shall  have 
a  given  length. 

465.  To  draw  a  straight  line  equidistant  from  three  given  points. 

466.  To  inscribe  a  straight  line  of  given  length  between  two  given 
circumferences  and  parallel  to  a  given  straight  line.     (See  Ex.  137.) 

467.  To  draw  through  a  given  point  a  straight  line  so  that  its  dis- 
tances from  two  other  given  points  shall  be  in  a  given  ratio  (m :  ri). 

HINT.   Divide  the  line  joining  the  two  other  points  in  the  given  ratio. 

468.  Construct  a  square  equivalent  to  the  sum  of  a  given  triangle 
and  a  given  parallelogram. 

469.  Construct  a  rectangle  having  the  difference   of  its   base   and 
altitude  equal  to  a  given  line,  and  its  area  equivalent  to  the  sum  of  a 
given  triangle  and  a  given  pentagon. 

470.  Construct  a  pentagon  similar  to  a  given  pentagon  and  equiva- 
lent to  a  given  trapezoid. 

471.  To  find  a  point  whose  distances  from  three  given  straight  lines 
shall  be  as  the  numbers  m,  n,  and  p. .  (See  Ex.  461.) 

472.  Given  two  circles  intersecting  at  the  point  A.     To  draw  through 
A  a  secant  BAC such  that  AB  shall  be  to  A C  in  a  given  ratio  (m  :  n). 

HINT.   Divide  the  line  of  centres  in  the  given  ratio. 

473.  To  construct  a  triangle,  given  its  angles  and  its  area. 

474.  To  construct  an  equilateral  triangle  having  a  given  area. 

475.  To  divide  a  given  triangle  into  two  equal  parts  by  a  line  drawn 
parallel  to  one  of  the  sides. 

*476.  Given  three  points  A,  B,  C.     To  find  a  fourth  point  Psuch  that 
the  areas  of  the  triangles  APB,  APC,  BPC,  shall  be  equal. 

477.  To  construct  a  triangle,  given  its  base,  the  ratio  of  the  other 
sides,  and  the  angle  included  by  them. 

478.  To  divide  a  given  circle  into  any  number  of  equivalent  parts  by 
concentric  circumferences. 

479.  In  a  given  equilateral  triangle,  to  inscribe  three  equal  circles 
tangent  to  each  other  and  to  the  sides  of  the  triangle. 


14  DAY  USE 

RN  TO  DESK  FROM  WHICH  BORROWED 


v  .*  •          "•!•?*/»* 

\  ,- 


This  book*  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 
Renewed  books  are  subject  to  immediate  recall. 


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. 


19S3 


APR  ]/4  1967 


19T1- 


LD  21-50m-6,'60 
(B1321slO)476 


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